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

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Answer 1 · 2015-12-25T03:59:31

$x = 14$ $x=14$

To undo a logarithm with base $3$ $3$ , exponentiate both sides with a base of $3$ $3$ .

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

$2 x - 1 = 27$ $2x-1=27$

$2 x = 28$ $2x=28$

$x = 14$ $x=14$

How do you solve log3(2x−1)=3log_3 (2x-1) = 3?