How do you solve #4e^(9x-1)=64#?

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Answer 1 · 2016-03-08T12:21:14

#x=(ln(16)+1)/9#

Divide both sides of the equation first. This will deal with the #4# attached to the #e^(9x-1)#.

#e^(9x-1)=16#

Now, take the natural logarithm of both sides of the equation.

#ln(e^(9x-1))=ln(16)#

Note that the #ln# and #e^x# functions are inverses, so they undo one another.

#9x-1=ln(16)#

Solving through basic algebra,

#x=(ln(16)+1)/9#