What is the value of $x$ in this logarithmic inequality?

Question

Please help me with this inequality :

$$\log_2 (x^2-2x) - 3 >0 $$

Answer 1

The logarithm in base 2 is a monotonically increasing function so you have that: $$\log_2 x > 3 \Leftrightarrow x > 2^3$$ In your case the argument of the logarithm is slightly more difficult but getting rid of the logarithm works in precisely the same way.

You'll end up with a quadratic inequality; can you take it from there?

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If you have trouble with understanding why you can get rid of the log in that way, here are two other ways of looking at it: $$\begin{array}{ll} \log_2 x > \color{blue}{3} \Leftrightarrow \log_2 x > \color{blue}{\log_2 8} \Leftrightarrow x > 8 & \quad \mbox{(rewrite $\color{blue}{3=\log_2 8}$)} \\ \log_2 x > 3 \Leftrightarrow 2^{\log_2 x} > 2^3 \Leftrightarrow x > 8 & \quad \mbox{(take both sides as the exponent of base $2$)} \end{array}$$