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#