How do you solve for x in #5^x=4^(x+1)#?

1 Answer
ProfLayton
Apr 21, 2016

#xapprox6.21#

Explanation:

First we'll take the #log# of both sides:
#log(5^x)=log(4^(x+1))#
Now there's a rule in logarithms which is: #log(a^b)=blog(a)#, saying that you can move any exponents down and out of the #log# sign. Applying this:
#xlog5=(x+1)log4#
Now just rearrange to get x on one side
#xlog5=xlog4+log4#
#xlog5-xlog4=log4#
#x(log5-log4)=log4#
#x=log4/(log5-log4)#
And if you type that into your calculator you'll get:
#xapprox6.21...#

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