two logarithm exercises asked by Falah Alnassri in الرياضيات في ذي قار والبصره

Exercise:

Solve in $\mathbb{R}$ , ${{\left( {{\log }_{2}}\left( -x \right) \right)}^{2}}-5{{\log }_{2}}{{x}^{2}}+25=0$

Solution: Let $t={{\log }_{2}}\left( -x \right)$ & $5{{\log }_{2}}{{x}^{2}}=10{{\log }_{2}}x=10{{\log }_{2}}\left| -x \right|$

So ${{t}^{2}}-10t+25=0\Leftrightarrow {{t}^{2}}-2\left( 5 \right)t+25=0\Leftrightarrow {{\left( t-5 \right)}^{2}}=0\Leftrightarrow t=5$

So ${{\log }_{2}}\left( -x \right)=5\Leftrightarrow {{\log }_{2}}\left( -x \right)=5{{\log }_{2}}2\Leftrightarrow -x={{2}^{5}}\Leftrightarrow x=-32$

Exercise:

Solve in $\mathbb{R}$ , ${{\log }_{4}}{{\log }_{2}}x+{{\log }_{2}}{{\log }_{4}}x=2$

Solution: Let $u={{\log }_{4}}x\Leftrightarrow u{{\log }_{4}}4={{\log }_{4}}x\Leftrightarrow {{4}^{u}}=x$

So ${{\log }_{4}}{{\log }_{2}}{{4}^{u}}+{{\log }_{2}}u=2\Leftrightarrow {{\log }_{4}}2u+{{\log }_{2}}u=2$

$\frac{\ln 2u}{\ln 4}+\frac{\ln u}{\ln 2}=2\Leftrightarrow \frac{\ln 2u}{2\ln 2}+\frac{\ln u}{\ln 2}=2\Leftrightarrow \frac{\ln 2}{2\ln 2}+\frac{\ln u}{2\ln 2}+\frac{\ln u}{\ln 2}=2$

$\Leftrightarrow \frac{1}{2}+\frac{\ln u}{\ln 2}\left( \frac{1}{2}+1 \right)=2\Leftrightarrow \frac{3}{2}{{\log }_{2}}u=\frac{3}{2}\Leftrightarrow {{\log }_{2}}u={{\log }_{2}}2\Leftrightarrow u=2$

Thus $2={{\log }_{4}}x\Leftrightarrow {{\log }_{4}}x=2{{\log }_{4}}4\Leftrightarrow x=16$