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# 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 $$k =$$$$-\Large\frac{1}{2}$$