How do you solve log_(2)x^2+log_(.5)x=5?

Feb 16, 2017

$x = 32$

Explanation:

${\log}_{2} {x}^{2} + {\log}_{0.5} x = 5$

Now ${\log}_{2} {x}^{2} = 2 {\log}_{2} x = \frac{2 \log x}{\log} 2$

and ${\log}_{0.5} x = \log \frac{x}{\log} 0.5 = \log \frac{x}{\log \left(\frac{1}{2}\right)} = \log \frac{x}{\log 1 - \log 2}$

= $- \log \frac{x}{\log} 2$

Hence ${\log}_{2} {x}^{2} + {\log}_{0.5} x = 5$

or $2 \log \frac{x}{\log} 2 - \log \frac{x}{\log} 2 = 5$

or $\log \frac{x}{\log} 2 = 5$

or $\log x = 5 \log 2 = \log {2}^{5} = \log 32$

Hence, $x = 32$