## 0568 Question 23

Forum for the GRE subject test in mathematics.
newb
Posts: 6
Joined: Wed Oct 29, 2008 4:02 am

### 0568 Question 23

For what value of b is the line y = 10x tangent to the curve y = e^bx at some point in the xy-plane?

1. 10/e
2. 10
3. 10e
4. e^10
5. e

I tried finding where the two eqns intersect.
I also tried taking the derivative of y = e^bx and setting that = 10

How should I approach this problem? I am finding it difficult to simplify 10/e when working backwords from the answer.

Thank you very much in advance.

Nameless
Posts: 128
Joined: Sun Aug 31, 2008 4:42 pm
use the form of the tangent line at x0

y=f'(x0)(x-x0)+f(x0)

We have f(x)= e^(bx) so f'(x)=be^(bx) then f'(x0)=be^(bx0)=10 (1)
Moreover since y=10x so we have f'(x0)x0=f(x0) (2)

from (1) &(2) you can solve for b=e/10

thmsrhn
Posts: 17
Joined: Fri Mar 26, 2010 7:18 am

### Re: 0568 Question 23

I didn t understand, Please try explaining one more time no?

mathQ
Posts: 41
Joined: Thu Mar 25, 2010 12:14 am

### Re: 0568 Question 23

At the point where line is tangent to curve, 10x = e^bx ------ (1)

at that point, dy/dx = b e^bx = 10 -------- (2)

from (1) put value of e^bx in (2)

so, b 10x = 10 ==> bx =1

put this in (1), so b = 10/e

elyd
Posts: 1
Joined: Sat Apr 13, 2013 6:03 pm

### Re: 0568 Question 23

Hi. I have a way that hopefully clears things up.

First, we set (i) 10x=e^bx because this is the point at which the two lines, y=10x and y=e^bx, will intersect in the xy-plane.

I solved for 10, and got (ii) 10=(e^bx)/x.

The derivative of y=e^bx is y'=be^bx. The derivative will equal 10 at a certain point, thus (iii)10=be^bx.

I set the two 10's equal to each other, thus (iv) (e^bx)/x=b(e^bx). If you solve for b on the right hand side, you get (v) b=1/x.

Plug (1/x) in for b in equation (i) and we find that x=e/10.

By subbing in e/10 for x in equation (v) it's clear that b=10/e.