Find x in this equation

Rearranging, we have

$\frac{1}{3} \times 9^{x+2}+\frac{1}{2} \times 9^{x+1} = 6 \times 4^{x+1} - 3 \times 4^x$

$9^{x+1} \left(\frac{1}{3} \times 9 + \frac{1}{2} \right) = 4^x \left( 6 \times 4 - 3 \right)$

$9^{x+1} \times \frac{7}{2} = 4^x \times 21$

$9^{x+1} = 4^x \times 6$

Rewriting in terms of powers of $2$ and $3$ , this is

$3^{2x+2} = 2^{2x} \times 6$

and finally $3^{2x+1}=2^{2x+1}$

The only way for both of these quantities to be equal is if the exponents on both sides are zero - ie if $2x+1=0$ ; so the required answer is $x=\boxed{-0.5}$ .

Used a spreadsheet to create two cells corresponding to the values of the left-hand side and right-hand side of the equation, as a function of one cell (x). Tried all five possibilities listed in the answer section. -0.5 is the value of x where the difference between the two functions is least of the five possibilities. The actual solution is approximately x = -0.4.

For x = -0.5, the value of the LHS is 9.171499 and the value of the RHS is 7.231885