How do you solve #4^x=6^(x+2)#?

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Answer 1 · 2016-03-24T07:33:40

In #4^x=6^(x+2)#, taking log of both the sides, we get

#x xx log4=(x+2)xxlog6# or

#x(log4-log6)=2log6# or

#x=(2log6)/(log4-log6)#

Now we can simplify using logarithmic tables.

As #log4=0.6021# and #log6=0.7782#

#x=(2xx0.7782)/(0.6021-0.7782)# or

#x=1.5564/-0.1761=-8.838#