Solving for the variable in an exponential function

[June]

Hey there,

I'm struggling trying to solve these questions. I've managed to figure out the exponential functions for both but don't know how to solve for r/t. Could someone please give me some guidance as to how to solve for the variables? Thank you.

 

[HallsofIvy]

That's what the logarithm is for! In fact, that is how the logarithm is defined. If $\displaystyle \frac{10000}{3}= (1.13)^t$ then, taking the logarithm of both sides, $\displaystyle \log\left(\frac{10000}{3}\right)= log(1.13^t)= t log(1.13)$. $\displaystyle x= \frac{\log\left(\frac{10000}{3}\right)}{log(1.13)}$.

Your calculator probably has two logarithm buttons- the "common" logarithm, base 10, and the "natural" logarithm, base e. As long as you use the same logarithm throughout a calculation, it doesn't matter which you use.

For $\displaystyle \left(1+ \frac{r}{4}\right)^{60}= \frac{1}{5}$, since the unknown, r, is NOT in the exponent itself, you do not need a logarithm. Instead, get rid of the exponent, 60, by doing the opposite of taking the 60th power- take the 60th root or 1/60 power:
$\displaystyle 1+ \frac{r}{4}= \left(\frac{1}{5}\right)^{1/60}$.

Now subtract 1 from both sides: $\displaystyle \frac{r}{4}= \left(\frac{1}{5}\right)^{1/60}- 1$.

And last, of course, multiply both sides by 4: $\displaystyle r= 4\left(\frac{1}{5}\right)^{1/60}- 4$.


That the exact solution and I, personally, would prefer it over an approximate, numerical solution. But it is not difficult to get that using a calculator. 1/5 is, of course, 0.2. The 60th root, or 1/60 power of 0.2 is, according to my calculator, 0.97353260205103889369927709540881. Multiplying by 4, that gives 3.8941304082041555747971083816352 and subtracting 4,
r= -0.10586959179584442520289161836478

 

[lookagain]

The negative r value is an error. That should have been a red flag, because the amount of money is growing.

The next two equations should be, after the first correctly written equation, the equivalent of:


$\displaystyle \bigg(1 \ + \ \dfrac{r}{4}\bigg)^{60} \ = \ \dfrac{10,000}{5,000}$

$\displaystyle \bigg(1 \ + \ \dfrac{r}{4}\bigg)^{60} \ = \ 2$

 

[Denis]

FV = (1 + i)^n : FV = Future Value

That represents future value of 1 unit;
really means FV = 1(1 + i)^n

 

[June]

Thanks for the help guys! Here's what I came up with (with your help of course) as for the final answer.