# Thread: Finding values of b

1. ## 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 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

Thank you

2. 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 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

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