exponentials help?

Question

Given 2^(x+1) + 2^x = 3^(y+2) - 3^y, where x and y are integers, what are the values of x and y

Answer 1

I assume that you want to just isolate y and isolate x. This then becomes two separate questions.

Let's first work on isolating y.

2^(x+1) + 2^x = 3^(y+2) - 3^y

Your first step is to note that a^m * a^n = a^(m+n). That is, the product of two numbers with the same base is just the base with the sum of their powers. In this case we'll be going in the opposite direction, noting that 2^(x+1) is the same as 2^x * 2^1.

(2^x)(2^1) + 2^x = (3^y)(3^2) - 3^y

Now, we're going to move the coefficients to the left of the number in question (as I'm trying to prove a point).

(2)(2^x) + 2^x = 9(3^y) - 3^y

Now, look at the above terms, and how they can be combined. See that 2(2^x) and 2^x? Think of the 2^x as a variable (say, y). Then we can combine them LIKE variables! The same way we add 2y + y to become 3y, we can treat the above JUST like variables, giving us

3(2^x) = 8(3^y)

So we want to solve for y, eh? The first thing you do is divide both sides by 8.

(3/8)(2^x) = 3^y

And then we convert this to logarithmic form.

log[base 3] [(3/8)(2^x)] = y

Similarly, to solve for x, relying on this equation we just solved for:
3(2^x) = 8(3^y)

Divide both sides by 3, to get

2^x = (8/3)(3^y)

And then convert to logarithmic form.

log [base 2] ( (8/3)(3^y) ) = x

If you're unsure of the steps to convert to logarithmic form, all you have to do is remember the following:
The BASE of the exponent *becomes* the BASE of the log. Everything else should be easy to memorize.

Good luck in your studies.

Answer 2

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