huda wrote:
If \(2^{8x}\)\(=640000\), then what is the value of \(2^{2x-2}\)?
A. \(5 \sqrt{2}\)
B. \(10\)
C. \(10 \sqrt{2}\)
D. \(80 \sqrt{2}\)
E. \(160\)
This is an intense question, mainly testing your knowledge of the rules of exponents.
The goal is to turn \(2^{8x}\) into \(2^{2x-2}\).First let's break down: \(2^{8x}\)\(=640000\), starting with 64,000.
\(2^{8x}\)\(=64,0000\)
\(2^{8x}\)\(=64*10,000\)
\(2^{8x}\)\(=2^6*10,000\)
\(2^{8x}\)\(=2^6*10^4\)
\(2^{8x}\)\(=2^6*(5*2)^4\)
Now, since we want \(2^{2x}\), let's break up \(2^{8x}\):
\((2^{2x})^{4}\)\(=2^6*(5*2)^4\)
To get rid of that power of four on the left side, let's quarter root both sides of the equation:
\((2^{2x})^{4*\frac{1}{4}}\)\(=2^{6*(\frac{1}{4})}*(5*2)^{4*(\frac{1}{4})}\)
We're left with:
\(2^{2x}\)\(=2^{\frac{6}{4}}*(5*2)\)
Now we need to get -2 in the exponent of \(2^{2x}\). Using the exponent rule of division with like bases, we can multiply both sides of the equation by \(\frac{1}{2}\) twice in order to accomplish that:
\(2^{2x}*\frac{1}{2}*\frac{1}{2}\)\(=2^{\frac{6}{4}}*(5*2)*\frac{1}{2}*\frac{1}{2}\)
\(2^{2x-2}\)\(=2^{\frac{6}{4}}*5*\frac{1}{2}\)
Simplifying the \(2^{\frac{6}{4}}\) we get:
\(2^{2x-2}\)\(=2^{\frac{3}{2}}*5*\frac{1}{2}\)
Now bringing that \(\frac{1}{2}\) into the \(2^{\frac{3}{2}}\) we get:
\(2^{2x-2}\)\(=2^{\frac{3}{2} - 1}*5\)
\(2^{2x-2}\)\(=2^{\frac{1}{2}}*5\)
Finally:
\(2^{2x-2}\)\(=5\sqrt{2}\)
So A is the answer