Using Logarithms to Compute Interest Rates

Logarithms are used in calculating investment yields based on interest rates.

Common calculations are computing how much money will be made on an investment of a given amount over a specified period at a given rate of interest that is compounded at a specified interval. A common question would be how much profit is made when \$1,000 is invested for three years at a four percent yield that compounded quarterly. The variables in this question are initial investment, profit, duration, interest rate, and compounding interval.

Other typical questions involving logarithms and interest rates are:

-    How long does it take to double an investment
-    Write the equation for computing how long it takes to double an investment
-    How long does it take for an investment to increase to a certain amount
-    Write an equation for simple interest
-    Profit after a given interval and interest rate if interest is compounded continuously
-    Profit after a given period of time with interest compounded at different rates
-    Given an interest rate and compounding interval then compute annual percentage yield
-    Given an initial investment amount, final amount, interest rate, and compounding interval compute the amount of time that it took for the investment to reach that amount
-    If an investment is now worth a given amount, and given a compounding interval, duration, and interest rate then what was the original investment amount

While some of these questions may seem wordy and confusing, they all consist of the same five variables and can be answered using four equations

The equation for computing yield on an investment that is compounded at a specific interval is

A = P(1 + r/n)^nt where:

• A is the current amount
• P is the initial amount invested, often called principal amount
• r is the investment rate such as five percent
• n is the number of times in a year that the interest is compounded
• t is time representing how long is the investment
• The equation for computing yield on an investment that is compounded continuously rather than at specified intervals is

A=Pe^rt where:

• A is the current amount
• P is the original principal amount
• e is Euler’s constant, approximately equal to 2.718 and computed as the limit as …
• r is the interest rate
• t is time or how long the money is invested
• An important question that will be answered in this section is how to write the equation for yield on an investment with continuous compounding using the equation for interest compounded at a specific interval.

The time to double an investment that is compounded continuously is ln(2)/r and if this looks familiar it should and it is because this is also the formula for computing half-life of a radioactive substance, which decays continuously. An important question is how this expression is computed.

The formula for computing simple interest is A = P(1+i)^k, where

• A is the current amount
• P is the principal amount
• i is the interest rate
• k is the number of times in a year that interest is compounded

An important question is how this expression is computed.

Example: Compute Annual Percentage Yield (APY).

Assume the interest rate is six percent and interest is compounded daily.

It is important to understand that if interest is compounded more than once a year that annual percentage yield (APY) will be greater than the interest rate.

The process for computing APY is to start with the standard expression for computing yield for an investment that is compounded at timed intervals:

A = P * (1 + r/n)^nt

In this case
A and P are not known
interest rate r is six percent or 0.06
time t is one year
n is the number of times in a year that interest is compounded. If interest is compounded daily then n is 365

The standard expression for computing yield is then

A = P * (1 + 0.06/365)^365*1
A = P (1.000164384)^365              1 + .06/365 = 1.000164384
A = P (1.061831311)                      1.000164384 ^ 365 = 1.061831311

The formula for simple interest, for one year, is A=P(1+r)^1. In this case (1 + r) = 1.061831311 so r = 1-1.061831311 = 0.061831311. This means annual percentage rate APY is 0.061831311 or 6.1831311 percent, or rounded to about 6.2 percent.

Note that because interest is compounded more than one time per year that APY is greater than interest rate. If interest were compounded once a year then APY would be equal to interest rate.

Example: Compute total amount.

If \$1,000 is invested at 12%, what is the total amount after three years if interest is compounded annually, semi-annually, quarterly, monthly, weekly, daily, and continuously. Place the results in a chart.

The place to start is with the standard expression for computing investment yield when interest is compounded at specific intervals:

A = P * (1 + r/n)^nt, where
A is unknown
P is principal or initial amount and is \$1,000.00 in this case
r is interest rate which is twelve percent or 0.12
t is time which is three in this case
n is compound frequency, which will be:
1 for annual
2 for semi-annual
4 for quarterly
12 for monthly
52 for weekly
365 for daily

Annual compounding: r=0.12, n=1, t=3

A = P * (1 + r/n)^nt
A = 1000 * (1 + 0.12/1)^1*3
A = 1000 * (1 + 0.12)^3
A = 1000 * (1.12)^3      Using a calculator
A = 1,404.93

Semi-annual compounding: r=0.12, n=2, t=3

A = P * (1 + r/n)^nt
A = 1000 * (1 + 0.12/2)^2*3
A = 1000 * (1 + 0.06)^6
A = 1000 * (1.06)^6      Using a calculator
A = 1,418.52

Quarterly compounding: r=0.12, n=4, t=3

A = P * (1 + r/n)^nt
A = 1000 * (1 + 0.12/4)^4*3
A = 1000 * (1 + 0.03)^12
A = 1000 * (1.03)^12      Using a calculator
A = 1,425.76

Monthly compounding: r=0.12, n=12, t=3

A = P * (1 + r/n)^nt.
A = 1000 * (1 + 0.12/12)^12*3
A = 1000 * (1 + 0.01)^36
A = 1000 * (1.01)^36      Using a calculator
A = 1,430.77

Daily compounding: r=0.12, n=365, t=3

A = P * (1 + r/n)^nt.
A = 1000 * (1 + 0.12/365)^365*3  Use a calculator here
A = 1,433.25

Note in this last example that no more intermediate results are computed because the results are numbers with a large number of digits and precision would be lost to rounding. It is better in a case like this to enter the full equation into a calculator (or math program like Jmathlib, matlab, or R) and round only the final result.

The formula for computing investment yield for continuously compounded interest is:

A = Pe^rt. In this case P = 1,000, e = Euler’s constant and this is a key on a calculator, r = 0.12 and t = 3

A = 1000 * e^(.12*3)
A = 1433.33

Note that more frequent compounding yields a greater return and that continuous compounding yields the best return.

Example: Develop continuous compound interest yield equation.

Using the formula for compound interest to develop an equation representing yield from continuous compounding.

The formula for periodic interest is

A = P (1 + r/n)^nt           and the formula for continuous interest is A=Pe^rt

The question is how is A=Pe^rt derived from A = P(1+r/n)^nt

A = P (1+ r/n)^nt

The key to this conversion is a variable substitution.
Let m = n/r.
With this, 1/m = 1/n/r = r/n and
n/t = m * rt = n/r * rt = nt

since 1/m = r/n the following substitution can be made:

A = P(1+1/m)^nt

Since nt = m*rt

A = P (1+1/m)^(m*r*t)
A = P [(1+1/m)^m]^rt     rules of exponents allows this

Look closely at the part that is (1+1/m)^m   The limit as m approaches infinity of (1+1/m)^m is Euler’s constant e. This allows for substituting (1+1/m)^m for e and the expression now is:

A = P * e ^ rt  and this is the expression for computing investment yields that have interest compounded continuously.

Example: Develop equation to compute time to double an investment.

Prove that time to double an investment can be computed using t = ln(2)/r and remember that half life is h = ln(2)/r if interest compounding is continuous

To prove that this expression works for computing time to double an investment start with the standard expression for computing yield from continuous compounding

A = P e^rt    A by definition is 2P so

2P = P e^rt   Dividing both sides by P leaves
2 = e^rt         Take natural log of both sides to drop the exponent down
ln(2) = ln e ^rt   Base e log of e to any power is that power
ln(2) = rt      isolate t by dividing both sides by r
t = ln(2)/r

Refer back to the process for computing how long is the half-life of a radioactive substance and the process is very similar

Example: Compute time to double an investment.

Compute time to double an investment if interest is compounded two times a year (semi-annually)

A = P(1+ r/n)^nt

If the final investment is doubled then A = 2P

2P = P(1+ r/n)^nt       Divide both sides by P and P goes away

2 = (1+r/n)^nt             take the log of both sides and the exponent can be brought down
ln(2) = ln(1+r/n)^nt
ln(2) = n*t*(ln(1+r/n)) the objective now is to isolate t
t = ln(2)/(n*ln(1+r/n))

If the initial investment is 5,000 and the compounding interval is 2 and the interest rate is 5% then what is the time to double the investment?

t = ln(2)/(n*ln(1+r/n))    n = t, r = 0.05, and the principal amount was previously eliminated from the expression.
t = ln(2)/(2*ln(1+0.05/2)) from a calculator
t ~= 14 years

Example: Computing simple interest.

If an amount of money P (for principal) is invested at an interest rate i for a period of time then after one time period the interest is P*i and the total amount if money called A is P + P*i.

In other words A = P + p*i

This equation can also be written as A = P(1 + i)

If the interest on the investment is reinvested then the new principal amount, which can be expressed as P + P*i or P *(1 + i), becomes P(1+i)(1+i) or P(1+i)^2

In general, after k number of periods the amount A is A=P(1+i)^k

Example: Compute amount after three years, time to grow to a fixed amount, and time to double

Assume the initial deposit is \$25,000, the interest rate is 5.4 percent per year, and interest is compounded daily.

Compute the amount after three years

A = P(1+r/n)^nt  In this case principal P, rate r, compounding interval n, and time t are known
P = 25000
r = 5.4 percent = 0.0054
n = 365 for daily compounding
t = three years
A is the unknown and the objective is to compute A

So the equation is:

P = 25000(1+.054/2)2*365

on a calculator this is 27850.97

This means that after three years the investment will grow from \$25,000 to  \$27,850.97

Compute time for investment to grow to \$35,000

A = P(1+r/n)^nt

In this case time is the unknown and all the other variables in this equation are known>

A is the final amount which is \$35,000
P is the principal amount which is \$25,000
interest rate r is 5.4 percent or 0.054
compounding interval n is 365 because interest is compounded daily

The objective is to solve for time t

After substituting in the known values the equation looks like this:

35000 = 25000 * (1 + 0.054/365)^365t

The first step is to divide both sides by 25000 making the 25000 on the right-hand side of the equation go away

7/5 = (1 + 0.054/365)^365t

The unknown variable t is in the exponent and in order to bring the exponent down, take the log of both sides

ln (7/5) = ln(1+ 0.054/365)^365t
ln (7/5) = 365*t*ln(1+ 0.054/365)

The exponent has been removed from the equation. Now it is possible to isolate time t by dividing both sides of the equation by 365 * ln(0.054/365)

t= ln(7/5)/(365*ln(1+0.054/365))

It is probably best to do variable substitution and calculation at the very end rather than compute intermediate results. Numbers can loose precision in rounding. It is best to do variable substitution at the end and plug the entire equation into a calculator or a computational program such as R or Matlab or JMathlib or one of many other such programs.

t = 6.23

This means a \$25,000 investment at 5.4 percent compounded daily will grow to \$35,000 in 6.23 years.

Compute how long it will take to double the investment to double from \$25,000 to \$50,000

To answer this question start with the standard equation for computing an investment with interest that is compounded at a timed interval:

A=P(1+r/n)^nt

It was proven in a prior example that the initial and final investment amounts are not required in the equation to answer the question and will be proven again here.

The final amount A is \$50,000
The initial amount P is \$25,000
The interest rate r is 5.4% or 0.054
The compound interval n is 365 because interest is compounded daily

Time is the unknown and the objective is to solve for time.

After substituting in the known values

50000 = 25000(1+.054/365)^(365*t)

The first step is to divide both sides by 25000. The result is

2 = (1+.054/365)^(365*t)

The next step is to take the logarithm of both sides. This moves the unknown value of t from the exponent to a value that can be algebraically manipulated and isolated.

ln(2) = 365*t*ln(1+.054/365)

Now t can be isolated by dividing both sides by 365*ln(1+.054/365)

t = ln(2)/(365*ln(1+.054/365))

Plugging this equation into a calculator or a math program such as R produces the result

t ~= 12.837

This means it will take about 12.837 years for the investment to double if the interest rate is 5.4 percent, and the compounding period is daily. It does not matter what the original principal amount is. For the given interest rate and compounding interval any and all investment amounts will double in this length of time.

As shown previously, the time to double a continuously compounded investment is ln(2)/r where r is the interest rate. If interest rate r is 0.054 then a continuously compounded investment will double in 12.836 years, very slightly less than an investment compounded daily.

Example: Compute amount after twelve years, time to grow to a fixed amount, and time to double.

Given an initial investment amount, or principal, of 12,000; interest rate of 6.3 percent, and a monthly compound interval, compute the amount after three years and how long it will take the account to grow to 20,000.

Compute the investment amount after three years:

To answer the first question about investment amount after three years, start with the standard equation for computing investment yield when interest is compounded at a fixed interval:

A = P(1+r/n)^(n*t)

Principal amount P is 12,000
Interest rate r is 6.3 percent of 0.063
Compounding interval n is 12

Substituting in the known values the equation now is

A = 12000*(1+0.063/12)^(12*3)

Plugging this equation into a calculator or a math program such as R, the result is:

A = 14489.33

Note that the process is to resolve the variables and compute the equation as the last step in the process. Doing it this way reduces the chance of error or loss of precision through rounding and helps to simplify the process.

How long will it take for the amount to grow from 12,000 to 20,000?

To answer this question, start with the standard equation for computing yield with interest compounded at fixed intervals:

A = P*(1+r/n)^(n*t)

In this case the final and principal amounts are known. The compounding interval n is known, and the interest rate r is known. The objective is to solve for t.

The final amount A is 20,000
Principal amount P is the original starting amount and is given as 12,000
The compounding interval n is 12
Interest rate r is 6.3 percent, or 0.063

Substituting in the known values, the equation now is:

20000 = 12000*(1+0.063/12)^(.063*t)

The unknown is time t and the objective is to solve for t.

As usual, the first step is to divide both sides by 12000 and simplify the result

20000/12000 = (1+0.63/12)^(12*t)

20000/12000 simplifies to 20/12 = 10/6 = 5/3

5/3 =  (1+0.63/12)^(12*t)

Time t is a value in the exponent so the next step is to bring down the exponent so that time t can be isolated. This is accomplished by taking the logarithm of both sides

ln(5/3) = ln((1+0.063/12)^(12*t)) the log of an exponent is the exponent …
ln(5/3) = (12 * t) * ln(1 + 0.063/12)
ln(5/3) = 12 * t * (ln(1 + 0.063/12))

The variable t can now be isolated by dividing both sides by (0.63 * ln(1+0.63/12)). This puts variable t on one side of the equation and all the numbers and calculations on the other side.

t = ln(5/3)/12 * (ln(1+0.063/12))

The time units were originally denominated in months so the result of the expression is in months. This means it will take about 154.8 months, or about 12.9 years, for the investment amount to grow from 12000 to 20000

t ~= 8.1 years

This would seem reasonable and a check can be made by computing the time to double an investment that is compounded continuously. That expression was previously proven to be ln(2)/r or in this case ln(2)/0.063 which is equal to 11 years. It would be expected to take less time for an investment to grow by 12000/20000, or 60 percent.

Example: If 9,500 is invested at 8.6 percent annually then compute the investment amount after three years if interested is compounded semi-annually, monthly, daily, and continuously.

To answer the question this time we will start with the computation for continuous compounding of interest. The reasons for this are that the equation is more simple, and continuous compounding will yield the largest final amount. For all other compound rates the result should be slightly less.

For continuous compounding the expression is

A = Pe^*rt) where

P is the original starting principal amount
e is Euler’s constant
r is the rate of interest
t is the amount of time

Filling in for the known variables the expression now is

A = 9500* e^(0.086*3)
Computing the result with a calculator or math program such as R the result is 12,296.22

To answer this question start with the standard equation for computing investment return when interest is compounded periodically.

A = P (1+r/n)^(n*t)

Principal amount P is \$9,500
interest rate r is 8.6% or 0.086
n is the compound frequency, which will first be 2 for semi-annually, then 12 for monthly, then 365 for daily.  A separate equation is used for computing continuously compounded interest.
time t is three years.

Substituting the variables into the equation the expression now is

A = 9500(1 + .086/2)^(2*3) for semi-annual compounding = 12,230.08
A = 9500(1 + .086/12)^(12*3) for monthly compounding    = 12,284.91
A = 9500(1 + .086/365)^(365*3) for daily compounding     = 12,295.85

The pattern of results is predictable
-     Continuous compounding yields a result greater than periodic compounding
-    The more frequent the compounding the greater the result

Example: If 5,420 is invested at 8.5% per year compounded quarterly then what is the investment amount after a year and a half?

A = P(1 + r/n)^(n*t)

Principal amount P is 5.420
interest rate r is 8.5% or 0.085
compound frequency n is 4 times per year
time t is 1.5 years

The objective is to solve for final amount A.

Substituting the known values into the equation gives

A = 5420 * (1 + .085/4)*^(4*1.5)

Entering this equation into a calculator or math program on a PC gives 6,148.82

This is a gain of about 729, or thirteen percent, over the original investment amount, over a 1.5 year period. Given the 8.5 percent annual rate the result appears reasonable.

After what period of time will the investment amount equal 7825?

To answer this question start with the standard equation for computing investment yield where interest is compounded at a fixed interval

A = P(1+r/n)^nt

In this case the amounts, compounding interval, and interest rate are known and the objective is to solve for time t.

Amount A is 7825
Principle amount P is 5420
compounding interval n is 4
interest rate r is 0.085

Substituting in the known values the equation now is

7825 = 5420 * (1 + 0.085/4)^(4*t)

The objective is to solve for time t which is in the exponent, which means there will be a step to take the logarithm of both sides.

The first step is to divide both sides by 5420 and reduce the fraction as much as possible

7825/5420 = (1 + 0.085/4)^(4*t)

The next step is to take the logarithm of both sides in order to make the exponent go away.

ln (7825/5420) = 4*t*(ln(1 + 0.085/4))

The next step is to divide both sides by 4*ln(1 + 0.085/4)) in order to isolate t

t = ln(7825/5420)/(4*ln(1 + 0.085/4))
t ~= 4.37

This means it will take about 4.4 years for the investment to increase by 2405 from 5420 to 7825.

If interest is compounded continuously how long will it take for the investment to increase to 7825?

The equation for computing with continuous interest compounding is

A = Pe^(rt)

Amount A is 7825
Principal, or original amount P is 5420
rate is 0.085

The objective is to solve for time t. The unknown value, time t, is once again in the exponent meaning that taking the logarithm of both sides will be used to isolate the variable in the exponent.

Rather than substituting in the values, the variables can be manipulated first to isolate time t.

The first step would be to divide both sides by P:

A/P = e^(rt)

The next step is to take the log of both sides.

ln(A/P) = ln(e^rt)

The base e log of e to a power is equal to the exponent, so

ln(A/P) = rt

Dividing both sides by r isolates time t

t = ln(A/P)/r

Substituting in the known values gives the expression

t = ln(7825/5420)/0.085

Using a calculator or a math program

t ~= 4.3 years

This means that with continuous interest that the investment will appreciate from 5420 to 7825 in 4.3 years.

Example: An account pays 5.2 percent annual interest that is compounded daily. If \$100,000 is invested how long will it take to accumulate 10,000 in interest.

To answer this question start with the equation for computing yield with interest computed at periodic intervals.

A = P (1+r/n)^(n*t)
If the account accumulates 10,000 in interest that means the total amount A is 110,000.
Principal amount P is 100,000

Interest rate r is 5.2% or 0.052
Compounding interval n is 365 because interest is compounded daily
time t is the unknown and the objective is to solve for t

The equation can be manipulated first to isolate the variable t and then after that values are substituted in. The unknown value t is in the exponent so plans should be made to take the log of both sides at the appropriate point

A=P(1+r/n)^nt

The first step would be to divide both sides by P to move that variable to the other side

A/P = (1+r/n)^nt

The next step is to take the logarithm of both sides in order to make the exponent go away.

ln(A/P) = n*t*ln(1+r/n)

The next step to isolate time t is to divide both sides by n*ln(1+r/n). The result is t isolated from all the known variables.

t = ln(A/P)/(n*ln(1+r/n))

The values for A, P, n, and r are known. Substituting in these values the equation is

t = ln(110000/100000) / (365*ln(1 + 0.052/365))
t ~= 1.8

This means it will take about 1.8 years for a \$100,000 investment to grow to \$110,000 at 5.2 percent interest that is compounded daily.

Example: A retirement plan pays 4.5 percent compounded continuously. Low long will it take to double in value.

This question requires use of the previously computed formula for computing time to double a continuously compounded investment:

t = ln(2)/r
t = ln(2)/0.045
t = 15.4

This means it will take 15.4 years for an investment that yields 4.5 percent with continuous compounding to double in value. Note that this is regardless of principal amount.

Example: Compute annual percentage yield (APY) if the interest rate is 4.5 percent compounded daily.

This is the same as computing amount A for one year:

A = P(1+r/n)^nt
A = P(1+0.045/365)^365*1
A = P(1.046)
APY=1.046 - 1 = 0.046 = 4.6 percent.

This means that the annual percentage yield is 4.34 percent.

If interest is compounded monthly:

A = P(1+r/n)^nt.
A = P(1+0.045/12)^12
A = P( 1.04594)
APY =  1.04594 - 1 = 0.04594 = 4.594 percent

This means if interest is compounded monthly the APY is 4.594 percent.

Example:  \$5,000 in an account pays 8.5 percent per year compounded quarterly.

Compute the amount in three years

The standard equation for computing interest compounded at a fixed interval is

A = P(1+r/n)^nt

P is the principal amount of 5000
r is the interest rate, 8.5% or 0.085
n is the compounding frequency which is 4 times per year
t is three years

The expression with variables substituted in is:

A = 5000(1+0.085/5)^4*3

On a calculator or using a math program this is 6,120.99

This means that in three years a 5,000 investment at 8.5 percent compounded quarterly grows to 6,120.99

Time to double an investment compounded periodically is

2P = P(1+r/n)^nt
2   =  (1+r/n)^nt
ln(2) = n*t*ln(1+r/n)
t = ln(2)/(n*ln(1+r/n))
t = ln(2)/(4*ln(1+0.085/4))
t = 8.24 years

Example: Compute the time required for a 5,420 investment at 7.5% compounded quarterly to increase to 7640.

A = P(1+r/n)^(nt)

Since time is the unknown variable it is best to first isolate the unknown variable and then replace the known variables with values

Note that time t is an exponent which means that at some point in the process a log of both sides of the equation will be taken in order to bring the variable down from the exponent.

First, divide both sides by principal amount P

A/P = (1+r/n)^(nt)

Next take the logarithm of both sides in order to bring down the exponent that has the unknown variable time

ln(A/P) = n*t*ln(1+r/n)

To isolate the variable t divide both sides by n * ln(1+r/n)

t = ln(A/P)/(n*ln(1+r/n))

Now that time t is isolated the next step is to substitute in the values for A, P, n, and r:

t = ln(7640/5420)/(4*ln(1+0.075/4))

Using a calculator or math program such as R on a PC the result is:

t ~= 4.62 years

This means that it takes 4.62 years for a 5,420 investment to grow by 2,220 to 7640 when interest is compounded quarterly at an annual rate of 7.5%.

Example: If \$4,823.94 is invested at 9.876 percent compounded semi-monthly, how long until 14793.04 is earned.

While this question appears annoyingly detailed the process is already established. Time t is the only unknown variable and the objective is to compute time based on certain values. It was seen in the previous example that t can be expressed as:

t = ln(A/P)/(n*ln(1+r/n))

In this case principal amount P is 4823.94
Total amount A is 4823.94 + 14793.04 = 19,616.98. A close look at this number means the principal will double twice at a relativey high interest rate so the expected result will be close to 15 years. ln(2)/r  is just over 7 years so twice that is the time to double twice.
rate r is 9.876% or 0.09876
the compounding interval is semi-monthly so n is 6

Substituting in the known values t is now equal to

t = ln(19616.98/4823.94)/(6*ln(1+0.09876/6))
t = 14.3 years

In examples such as this it is advisable to estimate the expected result so that when a number appears from R, Matlab, or on a calculator it can be evaluated as to whether the result is reasonable or possibly a mistake.

Example: Compute the interest rate. If \$1,000 is invested for four years and compounded twice a year. The amount has grown to \$1,435.77. Determine the interest rate.

Since interest is compounded at a fixed interval the equation to start with is the standard equation for computing yield on an investment that earns interest at a fixed interval.

A=P(1+r/n)^(nt)

Substituting in the known variables gives:

1435.77 = 1000(1 r/2)^(2*4)

The first time trying to solve this problem I tried to isolate the unknown variable without doing any substitution and the resulting expression became more complex and not simple. So it would appear the best approach may to be starting with substitution, then some computation, and see if things simplify.

Divide both sides by 1,000

1.43577 = (1+r/2)^8

Take the log of both sides to eliminate the exponent

ln(1.43577) = ln((1+r/2)^8)
ln(1.43577) = 8 * ln(1+r/2)

Divide both sides by 1/8 ti further isolate r

1/8 * ln(1.43577) = ln(1+r/2)
now make both sides an exponent of e to get r out of the log

e^(1/8 * ln(1.43577) = e^(ln(1+r/2))

e^(1/8 * ln(1.43577) = 1+r/2    subtract 1 and then double both sides to completely isolate r

r = 2 * ((e^(1/8 * ln(1.43577)) - 1)

r = 0.0925

This answer is reasonable for reasons as follows.

In the original question, the investment increased by nearly 50 percent in 4 years. It should double therefore in under eight. Time to double, if continuous, is ln(2)/r = ln(2)/0.0925 ~= 7.5 years. This means the answer is in the right ‘ball park’.

The answer to this question is saying that 1,000 invested at 9.25% compounded twice a year will appreciate to 1,435.77 in four years.

This example also shows that exclusive variable isolation before variable substitution is complex and not easily workable. A careful substitution of some variables, without making computations that cause loss to rounding, makes the problem easier to solve.

In this case interest rate r is a factor in an exponent which means the log of an equation will be taken.

First, divide both sides by P

A/P = (1+r/n)^(nt)

Next take the log of both sides to make the exponent go away

ln(A/P) = nt*ln(1+r/n)

start to isolate r by dividing both sides buy n*t

ln(A/P)/nt = ln(1+r/n)

Now the problem is the target variable is locked inside a logarithmic expression. The process of taking it out is to exponentiate both sides by the log base, which is e in this case.

e^(ln(A/P)/nt) = e^(ln(1+r/n))    e to the power of base e log is the power so

e^(ln(A/P)/nt) = 1+r/n               To finish isolating r subtract 1 from both sides and multiply times n

n * (e^(ln(A/P)/nt) -1) = r

The variables are:

n is 2 times per year compounding
A is 1435.77
P = 1000
t  is four years

So the expression with variables substituted in is

r = n * (e^(ln(A/P)/nt) -1)
r = 2 * (e^(ln(1435.77/1000)/2*4) - 1)

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