Question

If

$log_{2}(x) + log_{4}(x) = log_{0.25}( \sqrt{6} )$
Find the Value of X

Please solve it​

Answer 1

$log_{2}(x) + log_{4}(x) = log_{0.25}( \sqrt{6} )$

then x 0.408

Answer 2

$\log_2x+\log_4x\ =\ \log_{0.25}\sqrt6\\ \\ \\ \dfrac{\log x}{\log2}+\dfrac{\log x}{\log4}\ =\ \dfrac{\log\sqrt6}{\log 0.25}\\ \\ \\ \dfrac{\log x}{\log 2}+\dfrac{\log x}{\log(2^2)}\ =\ \dfrac{\log\left(6^{^{\frac{1}{2}}}\right)}{\log(2^{^{-2}})}\\ \\ \\ \dfrac{\log x}{\log2}+\dfrac{\log x}{2\log2}\ =\ \dfrac{\frac{1}{2}\log6}{-2\log2}}\\ \\ \\ \dfrac{-2\log x-\log x}{-2\log2}\ =\ \dfrac{\frac{1}{2}\log6}{-2\log2}\\ \\ \\ -3\log x\ =\ \dfrac{1}{2}\log6$

$\log x=\dfrac{\frac{1}{2}\log6}{-3}\\ \\ \\ \log x=-\dfrac{1}{6}\log 6\\ \\ \\ \log x=\log\left(6^{^{-\frac{1}{6}}}\right)\\ \\ \\ \log x=\log\left(\sqrt[6]{\dfrac{1}{6}}\right)\\ \\ \\ \\ \large\boxed{\therefore\ \ x=\sqrt[6]{\dfrac{1}{6}}\ \approx\ 0.742}$