Solve algebraically the equation 2 log5 x 3 log5 8 log5 2

Solve algebraically the equation 2 log_5 (x - 3) - log_5 (8) = log_5 (2)

Solution

2log5(x-3) - log5(8) = log5(2)

Use log property : loga -logb = log(a/b)

log5(x-3)^2 - log5(8) = log5(2)

log5(x-3)^2/8 = log5(2)

(x-3)^2/8 = 2

(x-3)^2 = 16

Take square root on both sides:

x-3 = +/-4

x = -1 , 7

With x= -1 term on original expression 2log5(-1-3) does not exist as for logx ; x>0

So, Only solution x = 7