How do you solve $92^x = 27^(x - 1)$ ?

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Answer 1 · 2016-11-14T18:51:35

Please see the explanation.

Take a logarithm of any base of both sides (I am going use the natural logarithm):

$ln(92^x) = ln(27^(x - 1))$

Use the property of all logarathims $log_b(a^c) = (c)log_b(a)$

$(x)ln(92) = (x - 1)ln(27)$

Use the distributive property:

$(x)ln(92) = (x)ln(27) - ln(27)$

Solve for x:

$(x)ln(92) - (x)ln(27) = - ln(27)$

$x = ln(27)/(ln(27) - ln(92))$

$x ~~ -2.68839$

How do you solve 92^x = 27^(x - 1)92^x = 27^(x - 1)?