Inequalities with powers and logarithms

Problem

Solve the following inequalities, using the properties of powers and logarithms where helpful: $a)\ \frac{4^{-2x}}{3^{-x}}>16\,;\qquad b)\ \big[\log_2 x\big]^{2}\le 16\,;\qquad c)\ \log_2\!\big(x^{2}\big)-\log_{1/2}(x)>3x.$

Solution

a) Rewrite $4^{-2x}=2^{-4x}=\left(\tfrac{1}{16}\right)^{x}$ and $\dfrac{1}{3^{-x}}=3^{x}$, so the left-hand side is $\left(\tfrac{1}{16}\right)^{x}3^{x}=\left(\tfrac{3}{16}\right)^{x}$. The inequality becomes $\left(\tfrac{3}{16}\right)^{x}>16$. Since the base $\tfrac{3}{16}<1$ the function is decreasing and the inequality reverses: $x<\log_{3/16}16=\frac{\ln 16}{\ln(3/16)}\approx -1{,}66.$ b) Let $t=\log_2 x$: $t^{2}\le 16\Rightarrow -4\le t\le 4\Rightarrow 2^{-4}\le x\le 2^{4}$, i.e. $\tfrac{1}{16}\le x\le 16$. c) With $x>0$ we have $\log_2(x^{2})=2\log_2 x$ and $\log_{1/2}x=-\log_2 x$, so the left-hand side is $2\log_2 x+\log_2 x=3\log_2 x$. The inequality becomes $3\log_2 x>3x$, i.e. $\log_2 x>x$: impossible, since $\log_2 x<x$ for every $x>0$. No solution. $\boxed{\,a)\ x<\log_{3/16}16\approx -1{,}66\,;\quad b)\ \tfrac{1}{16}\le x\le 16\,;\quad c)\ \nexists\,x\,}$