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

[solve]

Homework Statement

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.."

Homework Equations

The Attempt at a Solution

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

Thanks.

 

[Mentallic]

Is

\frac{1}{2}e^x&gt;e^x ??

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

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

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

 

[solve]

Is

\frac{1}{2}e^x&gt;e^x ??

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

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

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

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.

 

[Mentallic]

Is e^xlna=lna*e^x ?

Yes, they're equivalent. I prefer to have the log function at the end though just because people can confuse lna* e^x 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)*e^x

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

Thank You.

Yep! And if a=e then lna=1 so e^x*lna = e^x 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 e^xlna < e^x we need to add that this is only true for 0<a<e

 

[solve]

Yes, they're equivalent. I prefer to have the log function at the end though just because people can confuse lna* e^x 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)*e^x

Yep! And if a=e then lna=1 so e^x*lna = e^x 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 e^xlna < e^x we need to add that this is only true for 0<a<e

Thank You, Mentallic. It made my day.

 

[Ray Vickson]

Homework Statement

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.."

Homework Equations

The Attempt at a Solution

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

Thanks.

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

 

[solve]

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

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

 

[Mentallic]

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

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

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)

Is e^xlna=lna*e^x ?

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 &lt; 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.

 

[solve]

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

\left(e^{\ln(a)}\right)^x

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

 

[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

 

[Mentallic]

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

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

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)

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

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

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.

 

[solve]

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

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

 

[Mentallic]

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

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