# Thread: Trouble with X intercept question

1. ## Trouble with X intercept question

Hi, I'm having some issues with this problem:

$f(x) = -2e^{5x}+6$
I need to solve for the x intercept of the graph of f.

I tried to do it like so:

$0 = -2e^{5x}+6$
$2e^{5x} = 6$
$\log(2e^{5x}) = \log(6)$

$5x\log(2e) = \log(6)$

$5x = \frac{\log(6)}{\log(2e)}$

$x = \frac{\frac{\log(6)}{\log(2e)}}{5}$

That gives me: .2116484012

Using my graphing calculator to get the intersection of:
$y1 = -2e^{5x}+6$
$y2 = 0$

Would someone mind explaining what I've done wrong? Thanks!

2. The x intercept is where the graph crosses the x axis.

In other words you need to find f(0)

3. Isn't f(0) = 4 ? Or, the y intercept?

4. Originally Posted by cb220
Isn't f(0) = 4 ? Or, the y intercept?
Yes f(0) = 4 which is the/an x-intercept which is what the question asked for

The y intercept is

$2e^{5x} = 6$

$e^{5x} = 3$

$5x = \ln(3)$

$x = \dfrac{1}{5} \ln(3) \approx 0.220$

-------------

You went wrong between these two steps: $\log(2e^{5x}) = \log(6)$ and

$5x\log(2e) = \log(6)$

You can't do that because the exponent is only on $e$. For that to work you'd need the exponent on the 2 and the e. Instead you'd have to split the logs into $\log(2) + 5x \log(e) = \log(6)$

$5x\log(e) = \log(6) - \log(2) = \log(3)$

$x = \dfrac{\log(3)}{5\log(e)} \approx 0.220$

edit 2: You can use the change of base rule to see that $\ln(3) = \dfrac{\log(3)}{\log(e)}$

5. Ah, I see how I went wrong with my exponent not being attached to both terms. Thanks!

edit: However I'm still a bit confused on the x vs y intercepts. Putting my answers into the webwork form and using 4 as the y intercept and 2.1972246 as the x intercept came back as correct. I thought f(0) is the y intercept because the y intercept is where x=0 ? Conversely the x intercept would be where y = 0. Am I missing/forgetting something with that?

6. Originally Posted by cb220
Ah, I see how I went wrong with my exponent not being attached to both terms. Thanks!

edit: However I'm still a bit confused on the x vs y intercepts. Putting my answers into the webwork form and using 4 as the y intercept and 2.1972246 as the x intercept came back as correct. I thought f(0) is the y intercept because the y intercept is where x=0 ? Conversely the x intercept would be where y = 0. Am I missing/forgetting something with that?
From what I gather you have the definition of intercept the wrong way around.

The x intercept is where the graph crosses or touches the x axis. Thus, by deifintion the x value of that point(s) must always be 0. On your graph this makes (0,4) a point.

The y intercept is where it crosses or touches the y axis and hence y=0 and you will have (x,0) which usually requires more effort to find.

To find the x-intercept you calculate $f(0)$ and to find the y intercept you find $f(x)=0$. I hope that's clearer

7. Originally Posted by e^(i*pi)
The x intercept is where the graph crosses or touches the x axis. Thus, by deifintion the x value of that point(s) must always be 0. On your graph this makes (0,4) a point.
Isn't (0,4) a point on the y axis? 0 over and 4 up?

8. Yes, you're right. Turns out I got the intercepts the wrong way round

4 would be the y intercept and 2.20 would be the x intercept