How do you solve #12^(x-4)=3^(x-2)#?

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Answer 1 · 2017-01-07T14:01:34

#x=5.585#

As #12^(x-4)=3^(x-2)#, taking log on both sides, we get

#(x-4)log12=(x-2)log3#

or #xlog12-4log12=xlog3-2log3#

or #xlog12-xlog3=4log12-2log3#

or #x(log12-log3)=log((12^4)/(3^2))#

or #xlog4=log(4^2*12^2)=2log48#

or #x=(2log48)/(log4)=(2xx1.6812)/0.6021=5.585#