Solving variables in exponents

[c.w.richardjr]

Hello,

My question stems from a thought that I had when I was working on Problem Solving question #166 in the Quantitative Review 2nd Edition OG book (the green book).

If m is an integer such that (-2)^2m = 2^(9-m), then m =

The answer is 3 because when -2 is squared, it becomes a positive number. Therefore, it is true that the two bases on either side of the equation are considered "like bases," so we can then solve for m by making the exponents of either side of the equation.

First off, if this logic is incorrect, please let me know.

Secondly, what happens if the equation is (-2)^3m = 2^(9-m)? By raising -2 to the 3m, -2 does not become a positive integer. Is there a way to simplify if this were the case?

I hope this makes sense. Thank you.

[RonPurewal]

Hello,

My question stems from a thought that I had when I was working on Problem Solving question #166 in the Quantitative Review 2nd Edition OG book (the green book).

If m is an integer such that (-2)^2m = 2^(9-m), then m =

The answer is 3 because when -2 is squared, it becomes a positive number. Therefore, it is true that the two bases on either side of the equation are considered "like bases," so we can then solve for m by making the exponents of either side of the equation.

that's basically the idea.

more generally, you can replace "squared" with "raised to an even power" (since the exponent 2m is some even power, not necessarily the 2nd power).
you may already be aware of that, but just making sure.

First off, if this logic is incorrect, please let me know.

Secondly, what happens if the equation is (-2)^3m = 2^(9-m)? By raising -2 to the 3m, -2 does not become a positive integer. Is there a way to simplify if this were the case?

well, if that's going to work, then it's only going to work if the left-hand side becomes positive (since the right-hand side is clearly not going to be a negative number anytime soon).

so, you can always do this:
* temporarily ignore the "-" on the left-hand side
* solve the equation with two positive bases
* test each solution you get in the original equation
* keep the ones that work.

the boldface step is the key here -- because, by pretending that the bases are positive, you're admitting the possibility of "fake solutions" that actually turn out to be negative when you restore the original signs to the equation. by plugging back into the original, you can sniff out the fakes.