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

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Answer 1 · 2016-11-03T16:17:17

Please see the explanation for steps leading to a solution.

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 · 2016-11-03T21:35:23

#x =8#

#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#