# Help with logs

#### cistudent

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 (Doh)

Many thanks

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

#### TKHunny

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?

#### 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.

#### sa-ri-ga-ma

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.

#### 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

#### sa-ri-ga-ma

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 $$\displaystyle log_{10}{5} = 0.6990$$

#### cistudent

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

#### Niall101

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

• cistudent

#### Prove It

MHF Helper
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

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

• cistudent

#### cistudent

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

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

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.