Solve for x if (6^2x-3)/(9^x+3)=25
Rewriting 6 as 2×3 and 9 as 3^2, we get:
$$\frac{(2\cdot3)^{2x-3}}{3^{2x+6}} = 25$$
Simplifying the left-hand side, we obtain:
$$\frac{2^{2x-3}\cdot3^{2x-3}}{3^{2x+6}} = 25$$
$$2^{2x-3}\cdot3^{-9} = 25$$
$$2^{2x-3} = 25\cdot3^9$$
Taking the logarithm base 2 of both sides:
$$(2x-3)\log_2 2 = \log_2 (25\cdot3^9)$$
Simplifying:
$$(2x-3)\cdot1 = \log_2 (25\cdot3^9)$$
$$2x-3 = \log_2 (25\cdot3^9)$$
$$2x = \log_2 (25\cdot3^9)+3$$
$$2x \approx 21.56$$
$$x \approx 10.78$$
Therefore, the solution is approximately x = 10.78.