Question 377497
{{{log(4,(x-4))}}}+ {{{log(4,(x+2))}}}= {{{log(7,(49))}}}
since the bases on the left side are same that is 4 we can combine them using the laws of logarithms
{{{log(4,((x-4)(x+2)))}}}= {{{log(7,(7^2))}}}

{{{log(4,((x-4)(x+2)))}}}=2 {{{log(7,(7))}}}
{{{log(4,((x-4)(x+2)))}}}= 2*1 (since {{{log(7,(7))}}}=1 as the logarithm of a number to the same base is 1)
{{{log(4,((x-4)(x+2)))}}}=2
(x-4)(x+2)= 4^2 = 16
x^2 -2x-8= 16
x^2-2x-8-16=0 
x^2-2x-24=0
(x-6)(x+4)=0
so we have x = 6 and x = -4