# Math Help - Calculus Proof

1. ## Calculus Proof

I am having trouble using calculus to prove the following

Show that: 0<a<b when 1/b < (ln b - ln a)/b-a < 1/a

for example i know that 0<5<8 1/8 < (ln 8 - ln 5)/8-5 < 1/5 but i have no idea how to prove this

2. Originally Posted by greentea101
I am having trouble using calculus to prove the following

Show that: 0<a<b when 1/b < (ln b - ln a)/b-a < 1/a

for example i know that 0<5<8 1/8 < (ln 8 - ln 5)/8-5 < 1/5 but i have no idea how to prove this

The wording is terrible: apparently you want to show that if $0 , then $\frac{1}{b}<\frac{\ln b-\ln a}{b-a}<\frac{1}{a}$...**sigh**

Ok, use Rolle's theorem for the derivable function $f(x)=\ln x$

Tonio

3. ## see if this helps

if

(1/b) < [(ln b - ln a)/(b-a)] < (1/a)

then

(1/b) < (1/a) -- by simply ignoring the middle expression of the inequality

thus,

b > a

* now we must show that 'a' is greater than 0.

if

b > a

then

ln b > ln a

thus,

[(ln b - ln a)/(b-a)] > 0

if

0 < [(ln b - ln a)/(b-a)] < (1/a)

then

(1/a) > 0

thus,

a > 0

and therefore,

0 < a < b

4. Originally Posted by DanielJackson
if

(1/b) < [(ln b - ln a)/(b-a)] < (1/a)

then

(1/b) < (1/a) -- by simply ignoring the middle expression of the inequality

thus,

b > a

What if b<0 and a>0 ? Then it would be come a>b

* now we must show that 'a' is greater than 0.

if

b > a

then

ln b > ln a

If a<0 then ln a is not defined, so here you have assumed that 0<a<b which is
the thing you are trying to prove.

thus,

[(ln b - ln a)/(b-a)] > 0

if

0 < [(ln b - ln a)/(b-a)] < (1/a)

then

(1/a) > 0

thus,

a > 0

and therefore,

0 < a < b
This is not this simple.
It looks like you could use the mean value theorem....but I dont know.