How do you solve 2log_5(x-2)=log_5 36?

2 Answers
Nov 3, 2016

Please see the explanation for steps leading to a solution.

Explanation:

Given:

2log_5(x - 2) = log_5(36)

Divide both sides by 2:

log_5(x - 2) = (1/2)log_5(36)

Use the property (c)log_b(a) = log_b(a^c)

log_5(x - 2) = log_5(36^(1/2))

Replace 36^(1/2) with 6:*

log_5(x - 2) = log_5(6)

*Please notice that, technically, 36^(1/2) = +-6 but the negative value would violate the domain of the logarithm.

Make the logarithms disappear by writing both sides as exponents to base.

5^(log_5(x - 2)) = 5^(log_5(6))

x - 2 = 6

x = 8

Nov 3, 2016

x =8

Explanation:

2log_5(x-2) = log_5 36" "larr use the log power law

log_5 (x-2)^2 = log_5 36

Note: If log A = log B hArr A=B

:.(x-2)^2 = 36" "larr find the square root

x-2 = sqrt36 = +-6" "larr reject -6

x-2 = 6

x =8