Answer to Question #223284 in Discrete Mathematics for Vanessam

Question

Solve the equation log_xe^2e = eInx-e

Answer 1

$2elne/lnx=elnx-e$

$2=ln^2x-lnx$

$k=lnx$

$k^2-k-2=0$

$k_1=\frac{1+\sqrt{1+8}}{2}=2$

$k_2=\frac{1-\sqrt{1+8}}{2}=-1$

$lnx_1=2$

$x_1=e^2$

$lnx_2=-1$

$x_2=1/e$