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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;

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  • $\begingroup$ What do you mean by a fuller answer? $\endgroup$ Jul 5 '14 at 18:28
  • $\begingroup$ 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 $x>4$? $\endgroup$ Jul 5 '14 at 18:44
  • $\begingroup$ @PeterWoolfitt I did not see your comment before I posted. I should of just given a hint as well :). $\endgroup$
    – Chinny84
    Jul 5 '14 at 18:46
  • $\begingroup$ The solution is: for $x<0.25$ and for $x>4$. Chinny84 only got the second one. $\endgroup$
    – rae306
    Jul 5 '14 at 18:46
  • $\begingroup$ @Chinny you are still more than welcome to $\endgroup$
    – Bak1139
    Jul 6 '14 at 10:49
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$$ \log_b(x) = \frac{\log_a(x)}{\log_a(b)} $$ 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}$

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    $\begingroup$ 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. $\endgroup$
    – Ted
    Jul 5 '14 at 18:48
  • $\begingroup$ Quite correct!! $\endgroup$
    – Chinny84
    Jul 5 '14 at 18:50
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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.

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