workout wrote:
\(8^c * \sqrt{8}\) = \(\frac{8^a}{8^b}\) then a =
A) \(b(\frac{1}{2} + c)\)
B) \(\frac{bc}{2}\)
C) \(\frac{b + c}{2}\)
D) \(2b + c\)
E) \(\frac{1}{2} + b + c\)
When you're dealing with exponent questions, it's usually helpful to get things to look as alike as possible. For this problem, remember that roots are just exponents: a square root is the same as the \(\frac{1}{2}\) power.
So first, rewrite the equation: \(8^c * 8^\frac{1}{2}\) = \(\frac{8^a}{8^b}\)
Next, let's get rid of the denominator by multiplying both sides by \(8^b\)
\(8^b*8^c*8^\frac{1}{2} = 8^a\)
Next, apply the power rules. When multiplying numbers with the same base, you can add the powers:
\(8^b*8^c*8^\frac{1}{2} = 8^{(b+c+1/2)}\)=\(8^a\)
Now, since the base is the same on both sides, just ignore it. \(b+c+\frac{1}{2}\) = a.
Answer: E