Respuesta :
Answer:
[tex]x=\frac{8}{\ln(9)-1}[/tex]
Step-by-step explanation:
I will make an assumption.
We are trying to solve:
[tex]9^x=e^{x+8}[/tex]
We first need to have the same base on both sides. So I'm going to write [tex]9^x[/tex] as [tex]e^{\ln(9^x)}[/tex] using that natural log (ln( )) and e^() are inverses.
So we are going to rewrite the equation:
[tex]e^{\ln(9^x)}=e^{x+8}[/tex]
So now that the bases are the same we can set the exponents equal like so:
[tex]\ln(9^x)=x+8[/tex]
Using power rule we can bring down the x in front of ln( ):
[tex]x\ln(9)=x+8[/tex]
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:
[tex]x\ln(9)-x=8[/tex]
Factor the left hand side like:
[tex]x(\ln(9)-1)=8[/tex]
Now divide both sides by what x is being multiplied by:
[tex]x=\frac{8}{\ln(9)-1}[/tex]
