Fix this equation? Help 7^x-2 = 5^x+2?

7^(x - 2) = 5^(x + 2)
log[7^(x - 2)] = log[5^(x + 2)]
(x - 2)log(7) = (x + 2)log(5)
xlog(7) - 2log(7) = xlog(5) + 2log(5)
xlog(7) - xlog(5) = 2log(7) + 2log(5)
x[log(7) - log(5)] = 2[log(7) + log(5)
x = 2[log(7 * 5)]/[log(7/5)]
x = 2[log(35)]/[log(7/5)]

7^(x-2)= 5^(x+2)
log 7^(x-2) = log5^(x+2)
(x-2)log7 = (x+2)log5

x = 2[log7+log5]/[log7-log5]=2log35/log1.4 = 0.4513

Will ASSUME that you mean :-
7^(x - 2) = 5^(x + 2)
(x - 2) log 7 = (x + 2) log 5
(log 7 - log 5) x = 2 log 5 + 2 log 7
x = 2 (log 5 + log 7) / (log 7 - log 5)
x = 3.86

is it like 7 ^(x-2) = 5^(x+2) ?
well either way you have to use logarithms...