Posted on Categories - GMAT QuantNo. of comments 0

Challenge Problem 1 2016

If \({5}^{k^2}\times{25}^{2k}\times{625} = 25\times {\sqrt[4]{5}}\), which of the following is a possible value of \(k\)?

(A) \(-1\)

(B) \(-\Large\frac{1}{2}\)

(C) \(0\)

(D) \(\Large\frac{1}{2}\)

(E) \(1\)

First, notice that all the bases are powers of 5 – this includes the root which has exponent \(\frac{1}{4}\) (if you didn’t know that 625 is a power of 5, you need some more practice on Math FACTor 1). So, we can fiddle with the bases until we get something nicer:

\({5}^{k^2}\times{(5^2)}^{2k}\times{5^4} = {5^2}\times {5^{\frac{1}{4}}}\)

Using our rules of exponents we get

\({5}^{(k^2 + 4k + 4)} = {5}^{(2 + {\frac{1}{4}})}\)

Now that the bases are the same on both sides of the equation, we can set the exponents equal to each other.

\(k^2 + 4k + 4 = 2 + \)\(\Large\frac{1}{4}\)

A quadratic! Hopefully, you saw that coming. If not, don’t sweat it; you’ll see that pattern again, and next time you’ll recognize it.

Better yet, it’s a perfect square quadratic:

\({(k + 2)}^2 = 2 + \)\(\Large\frac{1}{4}\)

Now we need a bit of ingenuity. Let’s get rid of the fraction on the right hand side.

\({(k + 2)}^2 = 2.25\)

Square root both sides to get

\(k + 2 = \pm\sqrt{2.25}\)

The right hand side is one of your friends incognito: \(15^2 = 225\), so \(1.5^2 = 2.25\), so \(\sqrt{2.25} = 1.5\).

Hey, no whining! It’s a CHALLENGE problem.

\(k = -2 \pm{1.5}\)

\(k = -3.5\) or \(k = \)\(-\Large\frac{1}{2}\)

The correct answer is B.