Answer to Question #233557 in Algebra for sesi

Question

solve the following equation.

2log x = log 2x

Answer 1

$2log x = log 2x$

$log x^2= log 2x$ since $(xlogy= logy^x)$

Since the bases of logarithm are the same, set the arguments equal.

$x^2=2x$

$x^2-2x=0$

$x(x-2)=0$

$x=0$

$x-2=0$

$x=2>0.$

Answer: $x=2.$