Re: Solve the equation
Chipset3600 said:
Hi guys, please help me solving this =
\[\sqrt{\log_{a}\sqrt[4]{ax}+\log_{x}\sqrt[4]{ax}}+\sqrt{\log_{a}\sqrt[4]{\frac{x}{a}}+\log_{x}\sqrt[4]{\frac{a}{x}}}=a\]
\frac{1}{2}\sqrt{\log_{a}(ax)+\log_{x}(ax)}+\frac{1}{2}\sqrt{\log_{a}(\frac{x}{a})+\log_{x}(\frac{a}{x})}=a
\frac{1}{2}\sqrt{\log_a (a)+\log_a (x)+\log_x (a)+\log_x (x)}+\frac{1}{2}\sqrt{\log_{a}(x)-\log_{a}(a)+\log_{x}(a)-\log_{x}(x)}=a
\frac{1}{2}\sqrt{\log_a (x)+\log_x (a)+2}+\frac{1}{2}\sqrt{\log_{a}(x)+\log_{x}(a)-2}=a
\frac{1}{2}\sqrt{\log_a (x)+\frac{\log_a (a)}{\log_a (x)}+2}+\frac{1}{2}\sqrt{\log_{a}(x)+\frac{\log_{a}(a)}{\log_a (x)}-2}=a
Now let $t = \log_a(x) $\sqrt{t+\frac{1}{t}+2}+\sqrt{t+\frac{1}{t}-2}=2a
\sqrt{\frac{t^2+1+2t}{t}}+\sqrt{\frac{t^2+1-2t}{t}}=2a
\sqrt{\frac{(t+1)^2}{t}}+\sqrt{\frac{(t-1)^2}{t}}=2a
\frac{|(t+1)|}{\sqrt{t}}+\frac{|(t-1)|}{\sqrt{t}}=2a
here we can remove the absolute value since we are choosing
$t=\log_a(x) $ and t must be positive this is for x>1 then t>0
\frac{(t+1)}{\sqrt{t}}+\frac{\pm (t-1)}{\sqrt{t}}=2a
for t>1
\frac{2t}{\sqrt{t}}=2a
2\sqrt{t}=2a since t can't be negative $t=a^2$
Now substitute back for $t=\log_a(x) $
$\log_a(x) =a^2 \, \, \Rightarrow \,\, x=a^{a^2}$
for 0<t<1
\frac{(t+1)}{\sqrt{t}}+\frac{-t+1}{\sqrt{t}}=2a
\frac{2}{\sqrt{t}}=2a
\sqrt{t}=\frac{1}{a} \, \, \Rightarrow \,\, t=\frac{1}{a^2}
$x=a^{\frac{1}{a^2}}$
I hope I am making no mistakes >>> (Happy)
