Question

Solve for $x$:

$\sqrt{{\log }_x(ax{)}^{1/5}+{\log }_a(ax{)}^{1/5}}+\sqrt{{\log }_a{\left(\frac{x}{a}\right)}^{1/5}+{\log }_x{\left(\frac{a}{x}\right)}^{1/5}}=a$

• A. $x=a^{\left(5/4\right)a^2}$
• B. $x=a^{\left(4/5\right)a^2}$
• C. $x=\frac{5}{4}{a}^2$
• D. $x^2=a^{\left(5/4\right)a^2}$

Answer 1

The correct option is A $x=a^{\left(5/4\right)a^2}$

${\log }_x{\left(\frac{a}{x}\right)}^{\frac{1}{5}}+{\log }_a{\left(\frac{x}{a}\right)}^{\frac{1}{5}}$
$=\frac{1}{5}[{\log }_x\left(a\right)-1+{\log }_a\left(x\right)-1],(\because \log a^x=x\log a)$
$=\frac{1}{5}[\frac{\ln \left(x\right)}{\ln \left(a\right)}+\frac{\ln \left(a\right)}{\ln \left(x\right)}-2],(\because {\log }_{b}a=\frac{\log a}{\log b})$
Let $\frac{\ln \left(x\right)}{\ln \left(a\right)}=t$
$=\frac{1}{5}[t+\frac{1}{t}-2]$
$=\frac{1}{5}{(\sqrt{t}-\frac{1}{\sqrt{t}})}^2$...(i)
And ${\log }_x{\left(ax\right)}^{\frac{1}{5}}+{\log }_a{\left(ax\right)}^{\frac{1}{5}}$
$=\frac{1}{5}[{\log }_x\left(a\right)+{\log }_a\left(x\right)+2]$
$=\frac{1}{5}{(\sqrt{t}+\frac{1}{\sqrt{t}})}^2$ ...(ii)
Substituting in the equation we get $\frac{1}{\sqrt{5}}[\sqrt{t}-\frac{1}{\sqrt{t}}+\sqrt{t}+\frac{1}{\sqrt{t}}]=a$
$2\sqrt{t}=\sqrt{5}a$
$4t=5a^2$
$4{\log }_a\left(x\right)=5a^2$
$x=a^{\frac{5}{4}{a}^2}$.