Two logarithmic equations

Statement

Solve, discussing the domain conditions:

• (a) $\log_2 x+\log_2(x-6)=4$
• (b) $\log_5(2x+3)=1+\log_5(x-2)$

Solution

(a) Domain $x>6$. $\log_2\!\big(x(x-6)\big)=4\Rightarrow x^2-6x=16\Rightarrow x^2-6x-16=0\Rightarrow x=8$ or $x=-2$. Only $x=8$ is admissible.

(b) Domain $x>2$. $\log_5\dfrac{2x+3}{x-2}=1\Rightarrow \dfrac{2x+3}{x-2}=5\Rightarrow 2x+3=5x-10\Rightarrow x=\dfrac{13}{3}\approx 4{.}33$.

$\boxed{x_a=8;\qquad x_b=\tfrac{13}{3}\approx 4{.}33}$