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9^x= e^x+8
Solve for x using logs

User DMin
by
7.6k points

1 Answer

3 votes

Answer:


x=(8)/(\ln(9)-1)

Explanation:

I will make an assumption.

We are trying to solve:


9^x=e^(x+8)

We first need to have the same base on both sides. So I'm going to write
9^x as
e^(\ln(9^x)) using that natural log (ln( )) and e^() are inverses.

So we are going to rewrite the equation:


e^(\ln(9^x))=e^(x+8)

So now that the bases are the same we can set the exponents equal like so:


\ln(9^x)=x+8

Using power rule we can bring down the x in front of ln( ):


x\ln(9)=x+8

Now we need to get our terms with x on one side and terms without x on the opposite side. To do this, we just need to subtract x on both sides in this case:


x\ln(9)-x=8

Factor the left hand side like:


x(\ln(9)-1)=8

Now divide both sides by what x is being multiplied by:


x=(8)/(\ln(9)-1)

User Prashant Shilimkar
by
6.4k points