Finding values of b

• Nov 29th 2010, 07:10 AM
mybrohshi5
Finding values of b
Not sure if this post should go here so if it would be better else where would a Mod please move this :D Thanks!

Here is the question i am working on and it seems simple enough but i just wanted a second opinion

For what value(s) of b is $\displaystyle b^a$ > ab for all real numbers a?

if b is > 2 then $\displaystyle b^a$ will be > ab

but

I also notice that if a = 1 then b can also = 1 so i am not sure what to do with that.

Any help on how to prove this problem would be great :D

Thank you
• Nov 29th 2010, 08:34 AM
tonio
Quote:

Originally Posted by mybrohshi5
Not sure if this post should go here so if it would be better else where would a Mod please move this :D Thanks!

Here is the question i am working on and it seems simple enough but i just wanted a second opinion

For what value(s) of b is $\displaystyle b^a$ > ab for all real numbers a?

if b is > 2 then $\displaystyle b^a$ will be > ab

but

I also notice that if a = 1 then b can also = 1 so i am not sure what to do with that.

Any help on how to prove this problem would be great :D

Thank you

It must be $\displaystyle b>0$ since $\displaystyle b<0\Longrightarrow b^a$ is not defined for many real values of $\displaystyle a$. If $\displaystyle b=0$ then the

inequality is always true insofar we define $\displaystyle 0^0=1$.

If $\displaystyle a\leq 0$ the inequality is trivially true, as the LHS is positive and the RHS is non positive, so we can assume $\displaystyle a>0$, then:

$\displaystyle b^a\geq ab\Longleftrightarrow a\ln b\geq \ln a+\ln b\Longrightarrow (a-1)\ln b\geq \ln a$ , and since the ineq. is true for

$\displaystyle a=1$ we can assume $\displaystyle 0<a\neq 1$ , so it must be that

1) $\displaystyle a>1\Longrightarrow \displaystyle{\ln b\geq \frac{\ln a}{a-1}\Longrightarrow b\geq a^\frac{1}{a-1}}$ .

But $\displaystyle \displaystyle{\lim\limits_{a\to 1^+}a^\frac{1}{a-1}=\lim\limits_{a\to 1^+}e^\frac{\ln a}{a-1}=e}$ , so it must be $\displaystyle b\geq e$ , and this value already covers the whole

case $\displaystyle a>1$ , as $\displaystyle \displaystyle{a^\frac{1}{a-1}}$ has a maximum at $\displaystyle a=2$

2) $\displaystyle 0<a<1\Longrightarrow$ we get that $\displaystyle \displaystyle{a^\frac{1}{a-1}\xrightarrow [a\to 0]{}\infty}$ , so in this case $\displaystyle b$ cannot be bounded by some value.

Tonio