Find the number of real solutions of the equation $2\log_2\log_2x+\log_{\frac{1}{2}}\log_2(2\sqrt{2}x)=1$
My approach :
Solution : Here right hand side is constant term so convert it into log of same base as L.H.S. therefore, $1$ can be written as $\log_2\log_24$
$\implies 2\log_2\log_2x+\log_{\frac{1}{2}}\log_2(2\sqrt{2}x)= \log_2\log_24$
$\implies \log_2\log_2x^2 -\log_{2}\log_2(2\sqrt{2}x)= \log_2\log_24$
$\implies \log_2 \frac{\log_2x^2}{\log_2(2\sqrt2x)}= \log_2\log_24$
$\implies \frac{\log_2x^2}{\log_2(2\sqrt2x)}= \log_24$
Please suggest whether is it the right approach... thanks...
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- see entry 11 in our MathJax guide). $\endgroup$ – Zev Chonoles Oct 11 '13 at 4:55