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
 A: $$\color{violet}{(3^x)(4^{2x+1})=6^{x+2}}$$
$$\color{indigo}{3^x\cdot 2^{2(2x+1)}=2^{x+2}\cdot 3^{x+2}}$$
$$\color{blue}{2^{2(2x+1)}\cdot 3^x=2^{x+2}\cdot 3^{x+2}}$$
$$\color{green}{\frac{2^{2(2x+1)}}{2^{x+2}}=\frac{3^{x+2}}{3^x}}$$
$$\color{brown}{2^{2(2x+1)-(x+2)}=3^{(x+2)-x}}$$
$$\color{orange}{2^{3x}=3^2}$$
$$\color{red}{8^x=9}$$
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,
$$\color{red}{x=\frac{2\ln 3}{3\ln 2}}$$
A: $$3^x\cdot4^{2x+1}=6^{x+2}$$
$$\implies 3^x\cdot16^x\cdot4=6^x\cdot6^2$$
$$\implies\frac{3^x\cdot16^x}{6^x}=9$$
$$\implies\bigg(\frac{16}{2}\bigg)^x=9$$
$$\implies8^x=9$$
Taking log on both sides
$$\ln8^x=\ln9$$
$$\implies x=\frac{\ln9}{\ln8}$$
A: $$3^x4^{2x+1}=6^{x+2}$$
$$3^x2^{4x+2}=6^{x+2}$$
$$6^x2^{3x+2}=6^{x+2}$$
$$2^{3x+2}=36$$
$$2^{3x}=9$$
$$8^x=9$$
$$x=\log_8{9}$$
$$x=\frac{\ln 9}{\ln 8}$$
A: Take the logarithm of both sides, you will end up with a first degree equation:
$$x (\ln 3+2\ln 4 - \ln 6)=-\ln 4 + 2 \ln6$$
$$x \ln 8=\ln9...$$
Edit: computational error fixed
A: $$Log9/Log8$$
Just take all the terms with power of x on the one side and the rest on the other side..use the fact that $4^{2x}=16^x$..and combine it with $3^x$..and just arrange the terms
