Why would I divide these two equations to solve for i? I have the following two equations representing a longer actuarial practice question. I properly set up the equations, but am stumped on how to solve them.  The book says to divide the first by the second to get i.  So I can do that, I'm just having trouble intuitively grasping why.  Any explanations to help me understand?  Any other alternate ways of solving for i here.  
$$(1+i)^5(1+2i)^5=3.09$$
$$(1+i)^5(1+2i)^{15}=13.62$$
 A: There are really two separate questions tangled up here:


*

*Why can I divide one equation by the other?

*Why should I divide one equation by the other?


There are lots of things that you can do to this system of equations that don't make it easier to solve.  You could take the logarithm of both equations, for example, but it would just make the problem worse.  You could try to expand the first expression by using the binomial theorem twice and then multiply one fifth-degree polynomial by the other, and then do the same to the second equation, and then...  Never mind.  It doesn't help and becomes a total mess.
The short answer to why it is a good idea to divide one equation by the other is:  Because it makes the problem easier.  When you divide one equation by the other the common factor of $(1+i)^5$ cancels out and you are left with an equation that can be solved by "unwinding".  Note that that technique won't always work; it works for this problem because there is a common factor in both equations, and because the parts that are not identical differ only by an exponent, so the resulting quotient can be reduced.
A: The point of dividing is to simplify the term on the left: the $(1+i)^5$ will cancel out and you'll be left with $(1+2i)^{10}$ instead of the product of two factors involving $i$. Now that you have that on the left, it's easy to isolate $i$. (Start by raising both sides to the power of $\frac{1}{10}$.)
Incidentally, it's more intuitive to me to divide the second by the first, so that you have an exponent of $10$ instead of $-10$.
A: I'd actually divide the second by the first, but all we are doing is using the division property of equality and dividing both sides by $3.09$ the only difference is we want to achieve something simpler on the left so we use the fact that $(1+i)^5(1+2i)^5=3.09$ to simplify things.
A: Making more general, you have $$(1+a i)^{n_1}(1+b i)^{n_2}=A$$ $$(1+a i)^{n_1}(1+b i)^{n_3}=B$$ So, $$\frac AB=\frac{(1+a i)^{n_1}(1+b i)^{n_2}}{(1+a i)^{n_1}(1+b i)^{n_3}}=\frac{(1+b i)^{n_2}}{(1+b i)^{n_3}}=(1+bi)^{n_2-n_3}$$ Take logarithms of both sides $$\log(\frac AB)=(n_2-n_3)\log(1+bi)$$ from which $$i=\frac{\left(\frac{A}{B}\right)^{\frac{1}{{n_2}-{n_3}}}-1}{b}$$ In the case where you know in avance that $bi$ is small with respect to $1$, you could approximate the solution using $$\log(\frac AB)\approx(n_2-n_3)bi$$ In the case of you example, this would give $i\approx 0.074$ for an exact solution $i\approx 0.080$
