Is it even possible to solve 248.83 = 100*(1+x)^5 for x ?

[Alligator]

Hello everyone,

I'm new to calculus. Could I ask if it is possible to solve this equation for x?

248.83=100*(1+x)^5

(It is a formula from a compounding interest calculator. I am trying to solve for the compounding interest rate, often referred to as the 'effective interest rate' when the Future Value (3200) is known.)

For reference/context:

x is the compounding interest rate
248.83 is the Final Value (ie the balance after 5 years of investment)
100 is the initial investment (ie $100)
1 is the number of compounding periods per year
... and the power of 5 is for 5 years.

As mentioned I don't know if it is even possible to isolate x on one side of the equation but knowing either way will be very helpful.

Thank you very much for your time!

 

[frank789]

yes its possible

 

[Denis]

HINT: if a^n = b then a = b^(1/n)

 

[JeffM]

Hello everyone,

I'm new to calculus. Could I ask if it is possible to solve this equation for x?

248.83=100*(1+x)^5

(It is a formula from a compounding interest calculator. I am trying to solve for the compounding interest rate, often referred to as the 'effective interest rate' when the Future Value (3200) is known.)

For reference/context:

x is the compounding interest rate
248.83 is the Final Value (ie the balance after 5 years of investment)
100 is the initial investment (ie $100)
1 is the number of compounding periods per year
... and the power of 5 is for 5 years.

As mentioned I don't know if it is even possible to isolate x on one side of the equation but knowing either way will be very helpful.

Thank you very much for your time!

It is straight-forward using logarithms.

\(\displaystyle 248.83 = 100(1 + x)^5 \implies 2.4883 = (1 + x)^5 \implies ln(2.4883) = ln\{(1+x)\}^5 \implies \\

ln(2.4883) = 5 * ln(1+x) \implies ln(1 + x) \dfrac{ln(2.4883)}{5} \implies \\

1 + x = e^{\{ln(2.4883) \div 5\}} \approx 1.2 \implies x \approx 0.2.\)

Now check.

\(\displaystyle 100(1.2^5) = 120(1.2^4) = 144(1.2^3) = 172.8(1.2^2) = 172.8(1.44) = 248.832.\)

Close enough for government work.

 

[Deleted member 4993]

It is straight-forward using logarithms.

\(\displaystyle 248.83 = 100(1 + x)^5 \implies 2.4883 = (1 + x)^5 \implies ln(2.4883) = ln\{(1+x)\}^5 \implies \\

ln(2.4883) = 5 * ln(1+x) \implies ln(1 + x) \dfrac{ln(2.4883)}{5} \implies \\

1 + x = e^{\{ln(2.4883) \div 5\}} \approx 1.2 \implies x \approx 0.2.\)

Now check.

\(\displaystyle 100(1.2^5) = 120(1.2^4) = 144(1.2^3) = 172.8(1.2^2) = 172.8(1.44) = 248.832.\)

Close enough for government work.

Another close enough approximation (for -1< x <1):

ln(1 +x) = ln(2.4883)/5

x - x²/2 = 0.1823

x² - 2x + 0.3646 = 0

x = 0.20288

Now check:

ln(1+0.20288) = 0.184719 ...... continue...