Exponential Equation $4\cdot7^{x+2}=9^{2x-3}$

Question

Let $4\cdot7^{x+2}=9^{2x-3}.$

I do not know how to solve for $x$.

Progress

Took logarithms, got $$4(x+2\log7)=(2x-3)\log9$$ $$(x+2)\log7=[(2x-3)\log9]/4$$

Answer 1

It is clarified that we are to solve $4(7^{x+2}) = 9^{2x-3}$. Taking $\log$ is the right process but you have to make sure you do it right.

Using $\log(ab) = \log(a) + \log(b)$, the L.H.S. should be $\log 4 + (x+2) \log 7$ but not $4(x + 2) \log 7$.

Hence we have

\begin{align} \log 4 + (x+2) \log 7 & = (2x - 3) \log 9 \\ (\log 7)x + (\log 4 + 2\log 7) & = (2\log 9) x - 3 \log 9 \\ (2\log 9- \log 7)x & = \log 4 + 2\log 7 + 3 \log 9 \\ x & = \frac{\log 4 + 2\log 7 + 3 \log 9}{2\log 9- \log 7} \end{align}

Answer 2

$$4 \times {7^{x + 2}} = {9^{2x - 3}} \Leftrightarrow 4 \times 49 \times {7^x} = \frac{{{{81}^x}}} {{729}} \Leftrightarrow {\left( {\frac{{81}} {7}} \right)^x} = 4 \times 49 \times 729 = 142884 \Leftrightarrow x = {\log _{81/7}}142884.$$