Carcass wrote:
If \(9^{(x - \frac{1}{2})} – 2^{(2x – 2)} = 4^x – 3^{(2x – 3)}\), then what is the value of x ?
A. 3/2
B. 3/4
C. 4/9
D. 2/5
E. 1/5
Given: \(9^{(x - \frac{1}{2})} – 2^{(2x – 2)} = 4^x – 3^{(2x – 3)}\)
Rearrange: \(9^{(x - \frac{1}{2})} + 3^{(2x – 3)} = 4^x + 2^{(2x – 2)}\)
Rewrite \(9\) and \(4\) to get: : \((3^2)^{(x - \frac{1}{2})} + 3^{(2x – 3)} = (2^2)^x + 2^{(2x – 2)}\)
Apply the Power of a Power law to get: \(3^{(2x - 1)} + 3^{(2x – 3)} = 2^{2x} + 2^{(2x – 2)}\)
Factor each side as follows: \(3^{(2x – 3)}(3^2 + 1) = 2^{(2x – 2)}(2^2 + 1)\)
Evaluate: \(3^{(2x – 3)}(10) = 2^{(2x – 2)}(5)\)
Divide both sides by \(10\) to get: \(3^{(2x – 3)} = \frac{2^{(2x – 2)}(5)}{10}\)
Simplify: \(3^{(2x – 3)} = \frac{2^{(2x – 2)}}{2}\)
Simplify: \(3^{(2x – 3)} = 2^{(2x – 3)}\)
Divide both sides by \(2^{(2x – 3)}\) to get: \(\frac{3^{(2x – 3)}}{2^{(2x – 3)}}=1\)
Simplify: \((\frac{3}{2})^{(2x – 3)}=1\)
So, it must be the case that \(2x-3 = 0\)
Solve to get: \(x = \frac{3}{2}\)
Answer: A
Cheers,
Brent