How do you solve ${log}_{3} (2 x + 1) = 2$ $log_3(2x+1)=2$ ?

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Answer 1 · 2016-10-27T19:24:41

$x = 4$ $x=4$

${log}_{3} (2 x + 1) = 2$ $log_3 (2x+1) = 2$

$3^{2} = 2 x + 1$ $3^2 = 2x+1$

$9 = 2 x + 1$ $9=2x+1$

$8 = 2 x$ $8=2x$

$x = 4$ $x=4$

How do you solve log3(2x+1)=2log_3(2x+1)=2?