Hello Brent, I dont understand how you simplified this $3^{(2x – 3)} = \frac{2^{(2x – 2)}}{2}$ to get $3^{(2x – 3)} = 2^{(2x – 3)}$

Pls expand. thanks

For the rule of exponents $\frac{a^m}{a^n}=a^{m-n}$

we do have

$\dfrac{2^(2x)}{\frac{2^2}{2}}$ becomes $\frac{2^{2x-2}}{2^1}$ $====> 2^{2x-2-1} =====> 2^{2x-3}$

Thanks Carcass. Got it

$9^{(x- \frac{1}{2})}- 2^{(2x-2)} = 4^x -3^{(2x-3)}$

Our aim is to equalize the bases so that we can compare the exponents. To that end we manipulate the equation.

We bring all the identical bases to one side of the equation.

$9^{(x- \frac{1}{2})} + 3^{(2x-3)} = 4^x + 2^{(2x-2)}$

$3^{2(\frac{2x-1}{2})} + 3^{(2x-3)} = 2^2x + 2^{(2x-2)}$

$3^{(2x-1)} + 3^{(2x-3)} = 2^2x + 2^{(2x-2)}$

$3^{2x} \times 3^{-1} + 3^{2x} \times 3^{-3} = 2^{2x} + 2^{2x} \times 2^{-2}$

$3^{2x}[3^{-1} + 3^{-3}] = 2^{2x}[1 + 2^{-2}]$

$3^{2x}[\frac{1}{3} + \frac{1}{27}] = 2^{2x}[1 + \frac{1}{4}]$

$3^{2x}[\frac{10}{27}] = 2^{2x}[\frac{5}{4}]$

$3^{2x} = 2^{2x}[\frac{5}{4}] \times [\frac{27}{10}]$

$3^{2x} = 2^{2x}[\frac{27}{8}]$

$\frac{3^{2x}}{27} = \frac{2^{2x}}{8}$

$\frac{3^{2x}}{3^3} = \frac{2^2x}{2^3}$

$3^{2x} \times 3^{-3} = 2^{2x} \times 2^{-3}$

$3^{2x-3} = 2^{2x-3}$

The only way this equation can hold true is when both the exponents are equal to zero. Then we get $1 = 1$.

$2x-3=0$

$2x=3$

$x=\frac{3}{2}$