What is the value of $x$ in $6^{x+9}=4^{x-3}$?

Question

So I decided to do some more math to see if I was still able to do fairly complex math and came up with the problem$$\text{Solve for }x\text{: }6^{x+9}=4^{x-3}$$which I thought that I might be able to do. Here is my attempt at solving the aforementioned equation:$\require{cancel}{\color{white}{\text{.}}}$ $$6^{x+9}=4^{x-3}$$$$x\ln(6)+\ln(6^9)=x\ln(4)+\ln(4^{-3})$$$$\ln(6)(x+\ln(6^8))=\ln(4)(x+\ln(4^{-4})$$$$\dfrac{x+8\ln(6)}{\ln4}=\dfrac{x-4\ln(2)}{\ln6}$$$$\dfrac{x+8\ln(2)\ln(3)}{2\ln(2)}=\dfrac{x-4\ln(2)}{\ln(2)\ln(3)}$$$$\dfrac{x}{2\ln(2)}+\dfrac{\color{white}{.}^4\cancel{8}\cancel{\ln(2)}\ln(3)}{\cancel{2\ln(2)}}=\dfrac{x}{\ln(2)\ln(3)}-\dfrac{4\cancel{\ln(2)}}{\cancel{\ln(2)}\ln(3)}$$$$(x-2\ln(2))+4\ln(3)=(x-\ln(2)\ln(3))-(4-\ln(3))$$$$x-2\ln(2)+\color{white}{.}^3\cancel{4}\ln(3)=x-\ln(2)\ln(3)-4+\cancel{\ln(3)}$$$$x-2\ln(2)+3\ln(3)=x-\ln(2)\ln(3)$$$$\dfrac{x+3\ln(3)}{\ln(2)}-\ln(2)=\dfrac x{\ln(2)}-\ln(3)$$$$\dfrac{x+3\ln(3)-2\ln(2)}{\ln(2)}=\dfrac{x-\ln(3)\ln(2)}{\ln(2)}$$$$\implies x+3\ln(3)-2\ln(2)=x-\ln(6)$$$$x+\ln(6.75)=x-\ln(6)$$$$x+\ln(40.5)=x$$$$\ln(40.5)\neq0$$$$\therefore6^{x+9}\text{ will never be equal to }4^{x-3}$$

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My question

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Is my solution correct, or what could I do to attain the correct solution/attain it more easily?

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Mistakes I might have made

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1. Simplifying the logarithms
2. The solution overall

Answer 1

Another approach that makes for less "headache" in dealing with the algebra is to separate the exponential factors from the constant factors first:

$$ 6^{x+9} \ \ = \ \ 4^{x-3} \ \ \Rightarrow \ \ 6^x·6^9 \ \ = \ \ 4^x·4^{-3} \ \ \Rightarrow \ \ \frac{4^x}{6^x} \ \ = \ \ 4^3·6^9 $$ $$ \Rightarrow \ \ \frac{2^x}{3^x} \ \ = \ \ (2^2)^3·(2·3)^9 \ \ = \ \ 2^6·2^9·3^9 \ \ = \ \ 2^{15}·3^9 \ \ . $$

Now "take the logarithm" of both sides of the equation:

$$ \Rightarrow \ \ x·\ln 2 \ - \ x·\ln 3 \ \ = \ \ 15·\ln 2 \ + \ 9·\ln 3 $$ $$ \Rightarrow \ \ x \ \ = \ \ \frac{15·\ln 2 \ + \ 9·\ln 3}{\ln 2 \ - \ \ln 3} \ \ \ \text{or} \ \ \ -3· \left( \ \frac{5·\ln 2 \ + \ 3·\ln 3}{\ln 3 \ - \ \ln 2} \ \right) \ \ \approx \ \ -50.028 \ \ . $$

[check: $ \ \ 6^{-50.028 \ + \ 9} \ \ \approx 1.1852·10^{-32} \ \ \ ; \ \ \ 4^{-50.028 \ - \ 3} \ \ \approx 1.1852·10^{-32} \ \ ] $

Since the base $ \ 6 \ $ is raised to a power $ \ x \ $ increased by $ \ 9 \ \ , \ $ while the smaller base $ \ 4 \ $ is raised to $ \ x \ $ diminished by $ \ 3 \ \ , \ $ it is reasonable to expect that $ \ x \ $ will need to be a (fairly large) negative number.

Answer 2

$(x+9)\ln 6=(x-3)\ln(4)$, so $x(\ln 6-\ln 4)=-3\ln 4-9\ln 6$ and $$x=\frac{-3\ln 4-9\ln 6}{\ln 6-\ln 4}.$$