<s>[INST] Solving $3^x \cdot 4^{2x+1}=6^{x+2}$ Find the exact value of $x$ for the equation $(3^x)(4^{2x+1})=6^{x+2}$
Give your answer in the form $\frac{\ln a}{\ln b}$ where a and b are integers.
I have tried using a substitution method, i.e. putting $2^2$ to be $y$, but I have ended up with a complicated equation that hasn't put me any closer to the solution in correct form. Any hints or guidance are much appreciated.
Thanks
 [/INST] $$\color{violet}{(3^x)(4^{2x+1})=6^{x+2}}$$
$$\color{indigo}{3^x\cdot 2^{2(2x+1)}=2^{x+2}\cdot 3^{x+2}}$$
$$\color{blue}{2^{2(2x+1)}\cdot 3^x=2^{x+2}\cdot 3^{x+2}}$$
$$\color{green}{\frac{2^{2(2x+1)}}{2^{x+2}}=\frac{3^{x+2}}{3^x}}$$
$$\color{brown}{2^{2(2x+1)-(x+2)}=3^{(x+2)-x}}$$
$$\color{orange}{2^{3x}=3^2}$$
$$\color{red}{8^x=9}$$
Taking natural logarithm on both sides, we get
$$\ln 8^x= \ln 9$$
$$\color{blue}{x=\frac{\ln 9}{\ln 8}}$$
To be a bit more precise i.e. simplifying a bit more we can have,
$$\color{red}{x=\frac{2\ln 3}{3\ln 2}}$$
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