How to show $\log_2 x \cdot \log_{0.25} x \cdot \log_{0.125} x \cdot \log_{16} x > \frac {2}3$?

Question

I was trying to solve $\log_2 x \cdot \log_{0.25} x \cdot \log_{0.125} x \cdot \log_{16} x > \frac {2}3$ and I keep getting a partial answer of $x>4$ though answer key suggests a more expanded answer...?

Also, if you downvote, explain why;

Answer 1

using logaritham laws and a little algebra we get from $\log_2 x \cdot \log_{0.25} x \cdot \log_{0.125} x \cdot \log_{16} x > \frac {2}3$ to $(log_2x)^4>16$ which is the same as $|log_2x|>|2|$ feom which two options arise: $log_2x>2$ and $-(log_2x)>2$. union of the solution of both ob them would give $x<0.25 \cup x>4$ and uniting that with the range of definition (aka $x>0$) would yield $0<x<0.25 \cup x>4$ whihc is also final answer.

Answer 2

$$ \log_b(x) = \frac{\log_a(x)}{\log_a(b)} $$ doing this and convert ot base 2 we find $$ \log_2x\log_{0.25}x\log_{0.125}x\log_{16}x =\frac{\left[\log_2(x)\right]^4}{\left(\log_21 - \log_24\right)\left(\log_21 - \log_28\right)\log_216} $$ $$ \log_21 = 0 $$ we obtain $$ \frac{\left[\log_2(x)\right]^4}{-2\cdot-3\cdot 4}>\frac{2}{3} $$ thus $$ \left[\log_2(x)\right]^4 > 16=2^4 $$ therefore $$ \vert \log_2(x)\vert > 2 \implies x>4 $$ as @Ted pointed out there are more solutions namely $ 0 < x < \frac{1}{4}$