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

1 Answer
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...