Question
Solve the inequality
$$\log_{0.5}(2^x-1)\cdot\log_{0.5}\left(2^{x-1}-\frac12\right)\le 2$$
My Working
Let
$$t=\log_{0.5}(2^x-1)$$
We have that
\begin{align} \log_{0.5}(2^x-1)\cdot\log_{0.5}\left(2^{x-1}-\frac12\right)&\le 2\\ \log_{0.5}(2^x-1)\cdot\log_{0.5}\left(\frac{2^x-1}{2}\right)&\le 2\\ \log_{0.5}(2^x-1)\cdot\left[\log_{0.5}(2^x-1)-\log_{0.5}\left(\frac12\right)\right]&\le 2\\ \log_{0.5}(2^x-1)\cdot\left[\log_{0.5}(2^x-1)-1\right]&\le 2\\ t\cdot(t-1)&\le 2\\ t^2-t-2&\le0\\ (t-2)(t+1)&\le0\\ -1\le t&\le 2 \end{align}
I substituted the value of $t$ back, but there is a problem. For example, I did this for the left part of the inequality:
\begin{align} \log_{0.5}(2^x-1)&\ge -1\\ 0.5^{\log_{0.5}(2^x-1)}&\ge 0.5^{-1}\quad (\text{Mistake?!})\\ 2^x-1&\ge2\\ 2^x&\ge3\\ x&\ge \log_2 3 \end{align}
For some reason, I think the second line (of the above) is wrong, as the rest seems fine. Could anyone please tell me why I got this wrong and what the correct answer is? Thank you!
The incorrect line is: $$ \log_{0.5}(2^x-1)\cdot\left[\log_{0.5}(2^x-1)-\log_{0.5}\left(\frac12\right)\right]\le 2 $$ This should be: $$ \log_{0.5}(2^x-1)\cdot\left[\log_{0.5}(2^x-1)+\log_{0.5}\left(\frac12\right)\right]\le 2 $$ Or equivalently, you could write: $$ \log_{0.5}(2^x-1)\cdot\left[\log_{0.5}(2^x-1)-\log_{0.5}\left(2\right)\right]\le 2 $$ This gives you: $$ (t+2)(t-1)\le 0 $$ $$ -2\le t\le 1 $$ Which gives us two conditions: $$0< x\le \log_25$$ and $$x\ge \log_2\left(\frac{3}{2}\right)$$ with a final answer of: $$\log_2\left(\frac{3}{2}\right)\le x\le \log_25$$