# How to solve this logarithmic equation with sum of exponential functions?

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...

• Have you tried raising both sides to the power of $2$? – mwt Jul 9 at 15:07
• Oh gosh, thanks. – Arthur R. Jul 9 at 15:09
• These are not equations, they are inequalities. – Bernard Massé Jul 9 at 16:10

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