Question

Log base 6 (2x+1) = log base 6 (x-3) + log base 6 (x+5)

Answer 1

We assume you want to find values of x that make the expression true.

Use the rules of logarithms to rewrite as a single log. Then take the antilog and solve the resulting quadratic.

$\log_{6}(2x+1)=\log_{6}(x-3)+\log_{6}(x+5)=\log_{6}((x-3)(x+5))\\\\2x+1=(x-3)(x+5)=x^{2}+2x-15\\\\0=x^{2}-16=(x-4)(x+4)$

The term $\log_{6}(x-3)$ is only defined for x > 3, so the only solution to this equation is x = 4.