Help with logs

May 2010
11
0
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

\(\displaystyle z=e^{x^2}{^/}{^2}\)
 
Aug 2007
3,171
860
USA
Well, let's learn one more thing...

Please show your work.

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?
 
May 2010
11
0
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.
 
Jun 2009
806
275
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.
 
May 2010
11
0
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
 
Jun 2009
806
275
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 \(\displaystyle log_{10}{5} = 0.6990\)
 
May 2010
11
0
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
 
Aug 2008
98
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
 
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Prove It

MHF Helper
Aug 2008
12,883
4,999
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).



Now, looking at your question.

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

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

\(\displaystyle x = \pm\sqrt{2\ln{z}}\).
 
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May 2010
11
0
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.