Question

Log(base)4(3x-7)=log(base)4(x-15)

Answer 1

The bases of the logarithms on either side of the equation are the same, which means that their arguments must be the same:

$\log_4(3x-7)=\log_4(x-15)\implies3x-7=x-15$

Put another way, we can write both sides as powers of 4, then simplify and eliminate the logarithms:

$\log_4(3x-7)=\log_4(x-15)\implies4^{\log_4(3x-7)}=4^{\log_4(x-15)}\implies3x-7=x-15$

Then solve for $x$:

$3x-7=x-15\implies2x=-8\implies x=-4$

Then we just have to check if this solution is valid. On the left side, we have

$\log_4(3(-4)-7)=\log_4(-15)$

but we can't take logarithms of negative numbers, so there is no solution, as long as we're using the real-valued logarithm, anyway; $x=-4$ is a solution if we use the complex-valued version.