tonebeeze wrote:

Can someone please walk me through the simplification steps on this problem. I got lost along the way. Thanks!

If \(2^{x+2} - 3(2^{x-1})=5/8(2^{16-x)\), what is the value of x?

a. -7

b. 6

c. 7

d. 13

e. 16

There are two general principles that you can apply to any similar problem (a problem involving numbers and powers on either side of an equation):

1) get prime bases everywhere. Your goal is to get the same base on each side of the equation, in which case the exponents must be equal

2) if you are adding or subtracting two (or more) terms with the same base, factor out the *smaller* power

We can apply principle 2) on the left side of the equation in the question above, though it isn't as straightforward as in many other questions. If you were to see, for example:

\(7^7 + 7^5\)

you'd normally factor out the smaller power, which is the 7^5 term:

\(7^7 + 7^5 = 7^5 (7^2 + 1) = 7^5(50) = (2)(5^2)(7^5)\)

Notice that to get the two terms in the brackets (the 7^2 + 1), we just reduce the exponents by 5, because we factored out 7^5.

In the question above, we have an unknown in our power, but the same principle applies. If you see something like:

\(2^{x+2} - 2^{x-1}\)

we can again factor out the smaller power, which is 2^(x-1), to get

\(2^{x-1} (2^3 - 1)\)

Notice we are just subtracting x-1 from each power to work out the exponents on the two terms in the brackets.

Now, I think the way you've written the question might be confusing to some, because it appears that the 2^{16 - x} on the right side of the equation is in the denominator of a fraction. That can't be the case; then the equation cannot possibly be true. So I'm guessing the question ought to look like this:

\(\begin{align}

2^{x+2} - 3(2^{x-1}) &= \left(\frac{5}{8} \right) (2^{16-x}) \\

2^{x-1} (2^3 - 3) &= \left( \frac{5}{2^3} \right) (2^{16 - x}) \\

(2^{x-1})(5) &= (5)(2^{13 - x}) \\

2^{x-1} &= 2^{13 - x} \\

x - 1 &= 13 - x \\

x &= 7

\end{align}\)

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