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I come across this logarithmic equation recently (solve for $x \in \mathbb{R}$) : $$ 2x \geq \log_2 \left( \frac{35}{3} \cdot 6^{x-1} - 2 \cdot 9^{x - \frac{1}{2}} \right)$$ With few quick changes, this equation can be rewritten as : $$ \ln \left( \frac{4}{3} \right) x + \ln 3 \geq \ln \left(\frac{35}{6} \cdot 2^x - 2 \cdot 3^x \right)$$ So, how do you handle the right part ? Factoring doesn't appear to be so trivial...

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  • $\begingroup$ Have you tried raising both sides to the power of $2$? $\endgroup$ – mwt Jul 9 at 15:07
  • $\begingroup$ Oh gosh, thanks. $\endgroup$ – Arthur R. Jul 9 at 15:09
  • $\begingroup$ These are not equations, they are inequalities. $\endgroup$ – Bernard Massé Jul 9 at 16:10
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We need to solve $$2^{2x}\geq\frac{35}{18}\cdot6^x-\frac{2}{3}\cdot9^x,$$ where $$ \frac{35}{18}\cdot6^x-\frac{2}{3}\cdot9^x>0$$ and after substitution $\left(\frac{3}{2}\right)^x=t$ we obtain a quadratic inequality: $$\frac{2}{3}t^2-\frac{35}{18}t+1\geq0.$$ Can you end it now?

I got the following answer. $$(-\infty,-1]\cup\left[2,\log_{\frac{3}{2}}\frac{35}{12}\right).$$

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    $\begingroup$ Okay, thanks a lot for the substitution trick, I couldn't find this one. Thanks! $\endgroup$ – Arthur R. Jul 11 at 11:47

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