Maths C3 - Logs and Exponentials... HELP???

Can someone explain why this is?...

$e^{x ln a} = a^x$

I can't seem to understand how

If you understand that $e^{ln a}=a$, then the rest is simply application of $(a^b)^c \equiv a^{bc}$, as $e^{x ln a} \equiv (e^{ln a})^x \equiv (a^1)^x \equiv a^x$. I hope this is clear

That's the thing, I dont think I quite understand $e^{lna} = a$

Let's consider the equation $e^a = b$. How would we find $a$ in terms of $b$?

By taking logs to both sides. Then on the LHS use the "log power rule" to bring the constant a to the front. So... a = ln(b)

So $e^{ln b}=b$, as you have proved. This, in essence, is due to the definition of the natural logarithm; natural logarithms gives the exponent to which a base, in this case $e$, has to be raised to, to produce the argument, in this case $a$ or $b$. Hence, $e^{ln a} = a$, for all real $a$, $a \neq 0$.

Aren't you in year 11

I still don't really understand

Indeed I am.



Which bit is the source of confusion?

That's the thing, I dont think I quite understand $e^{lna} = a$

An approach that may make sense to you:

$\ln a \times \ln e = \ln a$

(true since $\ln e = 1$)

Using the log power rule as you call it:

$\ln e^{\ln a} = \ln a$

Then "cancelling out the logs":

$\Rightarrow e^{\ln a} = a$

Thank you, having yet another perspective has managed to clear this up I think

But how about the fact that...
$e^{x ln a} = a^x$ ?

This was explained earlier.

By the log power rule:

$e^{x \ln a} = e^{\ln a^x}$

If you're now happy that $e^{\ln{a}} = a$ for any $a$ then

$e^{\ln a^x}= a^x$

That's the thing, I dont think I quite understand $e^{lna} = a$

I didn't have time earlier but I'd like to give a more intuitive reason why this is true in addition to the log laws proof I gave earlier:

E.g. consider $\displaystyle 2^{\left(\log_2 8\right)}$

A logarithm is the power you need to raise a base by to get some number.

So $\log_2 8$ is the power you need to raise the base 2 by to get 8. Obviously this is 3.

Then $2^{\left(\log_2 8\right)}$ means "2 to the power of the thing you need to raise 2 by to get 8".

If you think about it, this is just going to take you back to 8 which means that $2^{\left(\log_2 8\right)}=8$. There is some "cancelling" going on here since $2^x$ and $\log_2 x$ are inverse functions.

Of course a similar principle will help you understand why $e^{\log_e x} = x$

I didn't have time earlier but I'd like to give a more intuitive reason why this is true in addition to the log laws proof I gave earlier:

E.g. consider $\displaystyle 2^{\left(\log_2 8\right)}$

A logarithm is the power you need to raise a base by to get some number.

So $\log_2 8$ is the power you need to raise the base 2 by to get 8. Obviously this is 3.

Then $2^{\left(\log_2 8\right)}$ means "2 to the power of the thing you need to raise 2 by to get 8".

If you think about it, this is just going to take you back to 8 which means that $2^{\left(\log_2 8\right)}=8$. There is some "cancelling" going on here since $2^x$ and $\log_2 x$ are inverse functions.

Of course a similar principle will help you understand why $e^{\log_e x} = x$

Ok so I may still be confused by this

You need to be much more specific, Im not sure if it was mentioned in a thread you posted or to someone else, but the more you attempt something and show your working, the more you get out of it.

If a teacher and a student were having a one-to-one and the student showed them a problem and they would say 'help me with this'.

The teacher would then say 'okay, which bit are you not understanding?'

It is a failing of the student if they then say 'I don't know / all of it' - it makes helping really difficult and what they get out of it less significant. A correct answer would be of the form 'I understand this and that but not sure about this bit' or 'I thought about using this method but wasn't sure how to apply it'.

I'm not having a go at you - just that you post many questions, which is fine, but my point is that you need to be much more specific if you're going to ask for help.


This is just saving the time of someone who would have replied to your post with 'which bit? / can you be more specific?'