How do I find zeros of an exponential function?

[GrannySmith]

I understand that normal exponential functions in the form f(x) = b^x has no x intercepts, but I'm also aware that when you translate the graph down any number of units or you reflect it over the x axis, it has x intercepts.

How would I find the value of these x intercepts for equations such as f(x) = (-e^x) + 4? I tried subtracting 4 then taking the natural log of each side but the natural log of 4 is undefined. I don't know how else to approach this problem.

 

[pka]

I understand that normal exponential functions in the form f(x) = b^x has no x intercepts, but I'm also aware that when you translate the graph down any number of units or you reflect it over the x axis, it has x intercepts.

How would I find the value of these x intercepts for equations such as f(x) = (-e^x) + 4?

First of all, there is no method (operations) used to work this problem. You look at it and see what works.

You need \(\displaystyle (-4)+4=0\) do you not?

So that means \(\displaystyle 4=e^x\), is that clear?

How do we now get \(\displaystyle x=\log(4)~?\) (you may use \(\displaystyle \ln\text{ for }\log\)).

 

[Steven G]

I understand that normal exponential functions in the form f(x) = b^x has no x intercepts, but I'm also aware that when you translate the graph down any number of units or you reflect it over the x axis, it has x intercepts.

How would I find the value of these x intercepts for equations such as f(x) = (-e^x) + 4? I tried subtracting 4 then taking the natural log of each side but the natural log of 4 is undefined. I don't know how else to approach this problem.

I can only think that the reason you feel that the equation becomes undefined is because e^x = -4?. Actually it is -e^x = -4 or e^x = 4. Then proceed as pka suggested.

 

[lookagain]

I understand that normal exponential functions in the form f(x) = b^x has [have] no x intercepts, but I'm also aware that when you translate the graph down any number of units > > > or you reflect it over the x axis, < < < it has x intercepts.

If the graph of f(x) = b^x is reflected across the x-axis, it will not cross the x-axis.

So it would not have any x-intercepts.