How do you solve $2^{x - 1} = 23$ $2^(x-1)=23$ ?

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Answer 1 · 2016-09-18T20:16:36

$:. x~~5.5236$ .

$2^(x-1)=23 rArr 2^x.2^-1=23 rArr 2^x=2*23=46$

Taking, $log_10$ of both sides,

$log_10 2^x=log_10 46$

$:. x*log_10 2=log_10 46$ .

$:. x=log_10 46/log_10 2=1.6628/0.3010~~5.5236$ .

How do you solve 2^(x-1)=232x−1=232^(x-1)=23?