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$$\log _{ 5 }{ (3^{ 2x }+6) } =\log _{ 5 }{ 3^{ x } } +1$$

Steps I took:

I realize that this actually means $5^{ \log _{ 5 }{ 3^{ x } } +1 }=3^{ 2x }+6$; however, I can't seem to find a clear cut path to the solution from here. I would prefer a hint that will help me arrive at the solution rather than the actual answer.

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  • $\begingroup$ Remember that $5^{a+b} = 5^a\times5^b$. $\endgroup$ – Paul Sinclair Oct 5 '15 at 3:59
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Just notice that $$\log _{ 5 }{ (3^{ 2x }+6) } =\log _{ 5 }{ 3^{ x } } +1\iff \log_5{(3^{2x}+6)}-\log_5 3^x = 1 \iff \frac{3^{2x}+6}{3^x}=5^1$$

Then, by putting $t=3^x$ you can solve last equation by solving $$t^2+6=5t$$ and after that, solving for $x$.

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Notice, we have $$\log_5(3^{2x}+6)=\log_53^{x}+1$$ $$\log_5(3^{2x}+6)=\log_53^{x}+\log_55$$ $$\log_5(3^{2x}+6)=\log_5(5\cdot 3^{x})$$$$3^{2x}+6=5\cdot 3^x$$ $$(3^x)^2-5(3^x)+6=0$$ $$(3^x-2)(3^x-3)=0$$ $$\implies 3^x-2=0\implies x\log 3=\log 2$$ $$\color{red}{x=\frac{\log 2}{\log 3}=\log_32}$$ $$\implies 3^x-3=0\implies x\log 3=\log 3$$ $$\color{red}{x=1}$$

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