The bases of the logarithms on either side of the equation are the same, which means that their arguments must be the same:
[tex]\log_4(3x-7)=\log_4(x-15)\implies3x-7=x-15[/tex]
Put another way, we can write both sides as powers of 4, then simplify and eliminate the logarithms:
[tex]\log_4(3x-7)=\log_4(x-15)\implies4^{\log_4(3x-7)}=4^{\log_4(x-15)}\implies3x-7=x-15[/tex]
Then solve for [tex]x[/tex]:
[tex]3x-7=x-15\implies2x=-8\implies x=-4[/tex]
Then we just have to check if this solution is valid. On the left side, we have
[tex]\log_4(3(-4)-7)=\log_4(-15)[/tex]
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; [tex]x=-4[/tex] is a solution if we use the complex-valued version.
