1. ## Help with logs

Could somebody please help a middle aged student who hasn't done any 'proper' maths for 25 years (and is starting to remember why he preferred English lessons) with this problem where I need to express with $\displaystyle x$ as the subject? I have the answer but mine doesn't match it

Many thanks

$\displaystyle z=e^{x^2}{^/}{^2}$

2. Well, let's learn one more thing...

You obviously know a logarithm is needed, so I'll give you this step.

$\displaystyle ln(z)\;=\;log_{e}(z)\;=\;\frac{x^{2}}{2}$

Now what?

3. OK so

$\displaystyle x=\sqrt2log_{e}z$

Which is the correct answer. How though did you get to that first stage? What am I not seeing - apart from the blindingly obvious, before you say it.

4. Originally Posted by cistudent
OK so

$\displaystyle x=\sqrt2log_{e}z$

Which is the correct answer. How though did you get to that first stage? What am I not seeing - apart from the blindingly obvious, before you say it.
If a^x = b then $\displaystyle x = \log_a{b}$

This formula is used to convert index form to log form.

5. I must be stupid. I'm still not getting it. Can you explain as if you were telling this to somebody who started his 'log chapter' today. Which I did, I hadn't even heard of $\displaystyle e$ before this afternoon.

Thanks

6. Originally Posted by cistudent
I must be stupid. I'm still not getting it. Can you explain as if you were telling this to somebody who started his 'log chapter' today. Which I did, I hadn't even heard of $\displaystyle e$ before this afternoon.

Thanks
OK. I hope you know $\displaystyle {a^m}\time{a^n} = a^(m+n)$
and \frac{a^m}{a^n} = $\displaystyle a^(m-n)$

10^0 = 1

10^1 = 10

Then 5 = 10^x, where value of x will be some where between 0 and 1.

In the calculator find log(5), you will see 0.6990.

That means 10^(0.6990) = 5.

It is written as $\displaystyle log_{10}{5} = 0.6990$

7. Nope! I'm still banging my head against the wall here. Could somebody take me through it step by step and explain each step. It's after 3.00am here and I can't sleep until I get this clear in my head

8. Take the log of both sides. log(e)^a = a

log and e^ and inverses of each other.

log(z) = log(e^(x^2/2))= (x^2)/2

9. By definition, a logarithm is the inverse of an exponential. So logarithms undo exponentials and exponentials undo logarithms.

In other words:

If $\displaystyle y = a^x$ then $\displaystyle \log_a{y} = x$ where $\displaystyle a$ is any base.

Proof: $\displaystyle y = a^x$

$\displaystyle \log_a{y} = \log_a{a^x}$

$\displaystyle \log_a{y} = x$ (since logarithms undo exponentials).

Similarly:

If $\displaystyle y = \log_a{x}$ then $\displaystyle a^y = x$ where $\displaystyle a$ is any base.

Proof: $\displaystyle y = \log_a{x}$

$\displaystyle a^y = a^{\log_a{x}}$

$\displaystyle a^y = x$ (since exponentials undo logarithms).

$\displaystyle z = e^{\frac{x^2}{2}}$.

In this case our base is the number $\displaystyle e$, Euler's Number. It pops up so often that you should research it...

We want to undo the exponential, so we use a logarithm of base $\displaystyle e$.

$\displaystyle z = e^{\frac{x^2}{2}}$

$\displaystyle \log_e{z} = \log_e{e^{\frac{x^2}{2}}}$

$\displaystyle \log_e{z} = \frac{x^2}{2}$ since the logarithm undoes the exponential

$\displaystyle 2\log_e{z} = x^2$

$\displaystyle x = \pm\sqrt{2\log_e{z}}$.

Note: Since the logarithm of base $\displaystyle e$ occurs so often in nature, it is called the Natural Logarithm, and is often denoted as $\displaystyle \ln$.

So it may be that the answer is written as

$\displaystyle x = \pm\sqrt{2\ln{z}}$.

10. Ah now I see. So $\displaystyle log_{e}e=1$

It's so bloomin obvious, why didn't I see that.

This was the last of a list of exercises and that whole $\displaystyle e^x{^2}{^/}{^2}$ thing threw me and made it seem more complicated than it really was, when all I had to do was bring down the whole exponent and multiply by one.

Thanks very much Niall and prove it. Now I can finally get to bed.