Quick Calculus question!

Loren Michael

So I have this problem, and I guess it's more of a logarithmic problem:

Find the area of the region (to two decimal places) bounded by the curve y = e^x and the ordinates y = 1 and y = 4 and the y-axis.

Now, given that this is dealing with the y-axis, I gotta do some switching around. My source suggests that I'm supposed to change the function to this prior to integrating: x = lny/lne.

I'm not sure what the rule is to get from y = e^x to there. I've been scouring my book and I can't justify why it's telling me to do that, or at least, I can't find how it gets there. Any help would be appreciated!

EDIT: Thanks people! Very helpful.

Starcross

Loren Michael wrote: »

So I have this problem, and I guess it's more of a logarithmic problem:

Find the area of the region (to two decimal places) bounded by the curve y = e^x and the ordinates y = 1 and y = 4 and the y-axis.

Now, given that this is dealing with the y-axis, I gotta do some switching around. My source suggests that I'm supposed to change the function to this prior to integrating: x = lny/lne.

I'm not sure what the rule is to get from y = e^x to there. I've been scouring my book and I can't justify why it's telling me to do that, or at least, I can't find how it gets there. Any help would be appreciated!

Take the natural log (ln) of both sides to get ln(y) = ln(e^x) = x.

ln(e)=1 so ln(y)/ln(e) is technically correct, as dividing by ln(e) won't do anything.

Edit: I'm not sure why it's got it written out in that way. Been a while since I've done any problems like this.

JoeUser

The important thing is to know the rules for dealing with e and ln (which is just log base e).

y = e^x

Apply ln to both sides

ln y = ln (e^x)

A rule of logs is that if you have something like log (x^5), you can rewrite that as 5*log(x). I'll explain more on that if you want.

So ln y = ln(e^x)
is

ln y = x*ln e

solve for x

x = ln y/ln e

And since ln e = 1, you have x = ln y