Solving equation with exponents

This is the original equation i started with and i am solving for $R$ $$2330 = \frac{10,000}{(1+R)^{25}}$$

After simplification i end up with the following:

$$(1+R)^{25} = 4.2918$$

How do i solve from here?I am trying to help a younger cousin with finance problems and she gets stuck on this kind of question. Any specific topics you guys can recommend that she can look at too brush up her algebra and log related equation solving skills?

• Take the 25th root on both sides and subtract 1 on both sides... – imranfat Feb 20 '14 at 19:58
• how will you do it on a calculator? i am sorry if this is a bad question to ask i am out of touch with this stuff l – user1772052 Feb 20 '14 at 20:00
• Type in $4.2918$ and raise it to the power $\frac{1}{25}$ Don't forget to put brackets on that "one over twentyfive" – imranfat Feb 20 '14 at 20:01
• Thanks a lot mate, so is there some website where one can review basic algebra etc. – user1772052 Feb 20 '14 at 20:09

When you raise an exponent to another exponent, the exponents can be multiplied:

$$(x^N)^M = x^\left(N*M\right)$$

So say you have an equation like:

$$x^N = a$$

If you raise both sides to the power 1/N:

$$(x^N)^\left(1/N\right) = a^\left(1/N\right)$$

That "cancels" the exponent on x:

$$x = a^\left(1/N\right)$$

Also, an exponent of $1/N$ is the same as the Nth root:

$$x = \sqrt[N]{a}$$

In your equation, after "cancelling" the exponent by taking both sides to the power 1/25, you still need to subtract 1 from both sides as well.

$$R = 4.2918^\left(1/25\right) -1$$

You received an exact answer to your problem from Bob Anderson and I shall try to provide you another one, more approximate, you could use to get easily an approximate solution you can use even with no calculator.

Your equation write $$f(R)=(1+R)^n-a =0$$ where you know that $R$ is small if compared to $1$. I shall use the fact that, for $R=0$, which should be our initial guess, all derivatives of $f(R)$ are very simple. The methods I shall use for finding the solution are what we use to call Newton type iterations (I shall not rewrite here the equations; you will find good descriptions at http://www.sztaki.hu/~bozoki/oktatas/nemlinearis/SebahGourdon-Newton.pdf).

Newton formula will give as estimate $R=\frac{a-1}{n}$
Halley formula will give as estimate $R=\frac{2 (a-1) n}{(a-1) (n-1) n+2 n^2}$

For your case, the exact solution is very close to $R=0.060$. The first order approximation gives $R=0.132$ while the second gives $R=0.051$ which is quite close (better than the order or magnitude). It could be possible to continue with higher order methods (such as Householder), but the formula will start to be too complex for hand manipulation.