# Y=e^xlna increases less quickly than y=e^x when a<e e. i. lna<1?

1. Feb 13, 2012

### solve

1. The problem statement, all variables and given/known data

I find this a bit tough to digest. Can somebody, please, elaborate?

"Because lna<1 when a<e you can see that the graph increases less quickly than the graph of e^x.."

2. Relevant equations
3. The attempt at a solution

Shouldn't e^xlna > e^x even if lna<1 ?

Thanks.

2. Feb 13, 2012

### Mentallic

Is

$$\frac{1}{2}e^x>e^x$$ ??

Of course not, because whatever ex is (we would determine its value if we were given x) half of it is always less since ex>0 for all x.

You can't extend this concept for all problems however. It wouldn't be completely true that $$x\ln(a)<x$$ for 0<a<e because if x = -2 for example, then half or any other value of ln(a), $-2\ln(a)>-2$

So keep in mind when doing these problems that you're dealing with a special case that ex>0 for all x.

3. Feb 13, 2012

### solve

Is e^xlna=lna*e^x ?

So, if a=1 and lna=0, then e^xlna<e^x, because e^xlna will equal 0 ???

Thank You.

4. Feb 13, 2012

### Mentallic

Yes, they're equivalent. I prefer to have the log function at the end though just because people can confuse lna* ex with $\ln(a*e^x) = \ln a+\ln (e^x) = \ln a + x$ so if you do it that way, just be sure to add parenthesis in the right places, mainly ln(a)*ex

Yep! And if a=e then lna=1 so ex*lna = ex in this case.
Also remember the rules of logs that you can only take the log of a number greater than zero, so when we say exlna < ex we need to add that this is only true for 0<a<e

5. Feb 13, 2012

### solve

Thank You, Mentallic. It made my day.

6. Feb 13, 2012

### Ray Vickson

Since you fail to use brackets, it is not clear what your expression e^xlna means. Is it $$e^{x \ln(a)} \mbox{ or } e^x \ln(a)?$$

RGV

Last edited: Feb 13, 2012
7. Feb 13, 2012

### solve

The first one. E to the power of X times natural log of A. Sorry for the ambiguity.

8. Feb 13, 2012

### Mentallic

Oh... This whole time I was going with the second one :yuck:

$$e^{x\ln(a)}$$ is equivalent to $$\left(e^{\ln(a)}\right)^x$$

edit: Something I have to note because I was going on the assumption that you were talking about $$e^x\cdot \ln(a)$$

No.

Sorry about getting the wrong ideas into your head. Keep in mind that everything I've been saying was intended to address the question of why $$\ln(a)\cdot e^x < e^x$$ for 0<a<e

The answer you're looking for however is in the hint I provided above about using the indice laws.

Last edited: Feb 13, 2012
9. Feb 13, 2012

### solve

$$\left(e^{\ln(a)}\right)^x$$ = a^x here?

10. Feb 14, 2012

### solve

Please, check to see if this is what I was trying to see from the get-go.

1. e^[x*ln(a)] <e^x if a<e

If a=2 and ln(a)=0.69, then

e^[x*ln(a)]: x=2, y=4 (approx)

e^x: x=2, y=8 (approx)

Here, y=e^x grows faster than y=e^[x*ln(a)]

2. e^[x*ln(a)] > e^x if a>e

If a=3 and ln(a)=1.1, then

e^[x*ln(a)]: x=4, y=81 (approx)

e^x: x=4, y=54(approx)

Here, y=e^[x*ln(a)] grows faster than y=e^x

11. Feb 14, 2012

### Mentallic

Yes, and since a<e, $$a^x<e^x$$ which is essentially all you need to understand.

y=4 exactly, not approximately. If a=2 then $$e^{x\ln(a)}=\left(e^{\ln(a)}\right)^x=a^x=2^2=4$$

Yes, pretty much. Using a few test values can help you get the idea of what is happening, but don't use it as an absolute proof.

12. Feb 14, 2012

### solve

Simple and elegant. This settles it. Thank You again, Mentallic.

13. Feb 14, 2012

### Mentallic

No worries, and sorry again about wasting your time and confusing you with all that nonsense before.