Answer to Question #223042 in Algebra for Darren Agbor

Question

Solve the equation log_xe^2e = eInx-e

Answer 1

$\log _{x} e^{2 e}=e \ln x-e \\\Rightarrow \log _{x} e^{2 e}=e \log _{e} x-e \\\Rightarrow \log _{x} e^{2 e}=\frac{e}{\log _{x} e}-e \\\Rightarrow 2 \mathrm{e} \cdot \log _{x} e=e\left(\frac{1}{\log _{x} e}-1\right) \\\Rightarrow 2 \log _{x} e=\left(\frac{1}{\log _{x} e}-1\right) \\Let \log _{x} e=t \\So, \\2 t=\frac{1}{t}-1 \\\Rightarrow 2 t^{2}+t-1=0 \\\Rightarrow 2 t^{2}+2 t-t-1=0 \\\Rightarrow 2 t(t+1)-1(t+1)=0 \\\Rightarrow(2 t-1)(t+1)=0 \\\Rightarrow 2 t-1=0\ or\ t+1=0 \\\Rightarrow t=\frac{1}{2},-1 \\As \log _{x} e=t$

We know that,

${ \log _{x} e>0}$

$So, \log _{x} e \neq-1 \\Now, \log _{x} e=\frac{1}{2} \\\Rightarrow e=x^{\frac{1}{2}}\\\Rightarrow x=e^2 \\Ans: x=e^2$