# Properties of logs question

1. Dec 3, 2008

### icystrike

1. The problem statement, all variables and given/known data
the question is this:
show
is 2$$\sqrt{3}$$

whereby e is natural logarithm base

2. Relevant equations

3. The attempt at a solution

Last edited: Dec 3, 2008
2. Dec 3, 2008

### lurflurf

Re: Logarithm

What have you tried?
Remember
x^y=e^[x log(y)]

3. Dec 3, 2008

### Integral

Staff Emeritus
Re: Logarithm

What do you know about the properties of logs?

4. Dec 3, 2008

### icystrike

Re: Logarithm

i suppose everything.. i just cant show tat it is 2sqrt3

5. Dec 3, 2008

### Integral

Staff Emeritus
Re: Logarithm

Do you have a text?

What are some of the properties of logs. They should be highlighted in boxes. RTFM

6. Dec 3, 2008

7. Dec 3, 2008

### Integral

Staff Emeritus
Re: Logarithm

look at the product rule:

Log(M*N) = logM + LogN

Now consider Log (M 2) = log (M*M) = log M + log M = 2log M

This is justification for the rule that is not shown on that page.

Log (MN) = N log M

Now apply that to the exponents of your problem.

8. Dec 3, 2008

### icystrike

Re: Logarithm

but it is just the exponent constant we cant apply the logarithm rules unless to the power.

9. Dec 3, 2008

### Integral

Staff Emeritus
Re: Logarithm

Repeat in english please

10. Dec 3, 2008

### icystrike

Re: Logarithm

oh.. let me rephrase
the equation is just the the sums of 2 exponent constant thus we cant apply the any logarithm rules except to the power of the constant

11. Dec 3, 2008

### Integral

Staff Emeritus
Re: Logarithm

You can apply the rules of logs to logs where ever they appear. Start by appling the rules to the logs which appear as exponents in your proplem.

$$e^{xlny} = e^{lny^x}$$

12. Dec 3, 2008

### icystrike

Re: Logarithm

yup i applied already
3e^(ln (1/sqrt3) ) + e^(ln sqrt3)

13. Dec 3, 2008

### HallsofIvy

Staff Emeritus
Re: Logarithm

You need: a ln(b)= ln(ba) and eln(x)= x.

14. Dec 3, 2008

### Integral

Staff Emeritus
Re: Logarithm

now you are getting some where.

you do know that :

$$e^{lnx} = x$$

15. Dec 3, 2008

### icystrike

Re: Logarithm

cool it works.
i have never seen this law before!
eln(x)= x

thank you HALLSOFIVY and INTEGRAL .
mind explaining this law eln(x)= x ?

16. Dec 3, 2008

### Integral

Staff Emeritus
Re: Logarithm

On the page YOU linked read the POWER RULE of logs.

then understand that ln = loge

17. Dec 3, 2008

### icystrike

Re: Logarithm

ya.. i know about it.
e^ln(x)= x

18. Dec 3, 2008

### Sjorris

Re: Logarithm

The natural logarithm is the 'inverse' of the e-exponential. If ex is f(x), and ln(x)=g(x), then you can rewrite the equation eln(x) to f(g(x)), but g(x)=f-1(x) (it's inverse), so f(f-1(x))=x. Or qualitatively, applying a function (the exponential) to it's inverse (the natural logarithm), or vice versa, returns it's argument ('input'), they 'undo' each other.

I'm sure there is a more rigorous and mathematically correct proof out there though.

19. Dec 3, 2008

### icystrike

Re: Logarithm

i do agree with your proving however until that inverse portion what do you really do to inverse it? i know it is somehow inverse of its law but how do you really come about to do it?

20. Dec 3, 2008

### icystrike

Re: Logarithm

btw you see, ln x means x=e^? and e^x means e x e x e x e x ...
^
l
x factors
i dont really see the relationship