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

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;

• What do you mean by a fuller answer? Jul 5 '14 at 18:28
• General Hint: Remember the logarithm identities $\log_ba=\dfrac{\log a}{\log b}$, and $\log a^b=b\log a$. Can you show us how you got to $x>4$? Jul 5 '14 at 18:44
• @PeterWoolfitt I did not see your comment before I posted. I should of just given a hint as well :). Jul 5 '14 at 18:46
• The solution is: for $x<0.25$ and for $x>4$. Chinny84 only got the second one. Jul 5 '14 at 18:46
• @Chinny you are still more than welcome to Jul 6 '14 at 10:49

$$\log_b(x) = \frac{\log_a(x)}{\log_a(b)}$$ 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 $0 < x < \frac{1}{4}$
• At the very end, after $[\log_2{x}]^4 = 2^4$, you should get $|\log_2{x}| > 2$ so there are more solutions where $\log_2{x}$ is negative.
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.