How do you solve #log_4 (3x+1) = 2#?

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Answer 1 · 2015-12-20T19:46:49

x = 5. I solved it by sight ... but how?

To solve any equation for x, you want to do the same thing to both sides until you isolate the x on the left side.

First undo the log expression by raising 4 to both sides:

#log_4(3x+1)=2#

#4^(log_4(3x+1))=4^2# and by the inverse property we get

#3x+1=16# which easily solves to #x=5.#

To check your answer put in #x=5# and get

#log_4(3(5)+1)=log_4(16)=2# because #4^2=16.#

/ dansmath strikes again! \