Question

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

Answer 1

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`

Answer 2

`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`