How do you solve ${log}_{4} (3 x + 1) = 2$ $log_4 (3x+1) = 2$ ?

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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:

${log}_{4} (3 x + 1) = 2$ $log_4(3x+1)=2$

$4^{{log}_{4} (3 x + 1)} = 4^{2}$ $4^(log_4(3x+1))=4^2$ and by the inverse property we get

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

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

${log}_{4} (3 (5) + 1) = {log}_{4} (16) = 2$ $log_4(3(5)+1)=log_4(16)=2$ because $4^{2} = 16 .$ $4^2=16.$

/ dansmath strikes again! \

How do you solve log4(3x+1)=2log_4 (3x+1) = 2?