Properties of Ln and e^x)

[Tricky557]

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

I'm just not sure what the answer to this is. I think it's an identity for e^x and ln, but I've never had a course that dealt with e^x or logs. So I don't know.

What is the answer to e^14ln(x)? It's part of a larger problem, but I can't get the rest of it done until I know that.


2. Relevant equations

None.

3. The attempt at a solution

I think the answer is x^14. But I'm not sure.

[Hurkyl]

Well, what was your reasoning?



Aside: when you're writing linearly, be careful about parentheses! The expression
e^14ln(x)
means
e^{14} \ln(x)
whereas
e^(14ln(x))
means
e^{14 \ln(x)}

[Tricky557]

Yes, sorry about that. I did mean to write:

e^(14*ln(x))

I was thinking, that if I equated some random variable(say y) to e^(14ln(x)), then I could just solve that equation.

y= e^(14*ln(x))
lny= 14lnx

lny = 14lnx

lny = ln(x^14)

e^ of both sides

y = x^14


The part of that I am unsure about is:

Does 14lnx = ln(x^14) ?

[daveb]

Yes, that works, so you can see that by definition, elnx = x. Also, a*lnx = ln(ax)

[Mark44]

Yes, sorry about that. I did mean to write:

e^(14*ln(x))

I was thinking, that if I equated some random variable(say y) to e^(14ln(x)), then I could just solve that equation.
That's not the best approach. The intended goal of this problem is for you to simplify the given expression. Setting your expression equal to, say, y, doesn't help much to move things toward your goal of simplification.

Use the properties of logs and exponents to rewrite your expression in a different (and simpler) form.


y= e^(14*ln(x))
lny= 14lnx

lny = 14lnx

lny = ln(x^14)

e^ of both sides

y = x^14


The part of that I am unsure about is:

Does 14lnx = ln(x^14) ?Yes.

[Tricky557]

Thanks for the help!