# Homework Help: Properties of logs question

1. Dec 3, 2008

### icystrike

1. The problem statement, all variables and given/known data
the question is this:
showhttp://img80.imageshack.us/img80/8628/qnszb4.jpg [Broken]
is 2$$\sqrt{3}$$

whereby e is natural logarithm base

2. Relevant equations

3. The attempt at a solution

Last edited by a moderator: May 3, 2017
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

### icystrike

Re: Logarithm

http://www.math.unc.edu/Faculty/mccombs/web/alg/classnotes/logs/logprops.html [Broken]

i think tats all of it?

Last edited by a moderator: May 3, 2017
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

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

3e^(ln (1/sqrt3) ) + e^(ln sqrt3)

13. Dec 3, 2008

### HallsofIvy

Re: Logarithm

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

Last edited by a moderator: May 3, 2017
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

then understand that ln = loge

17. Dec 3, 2008

### icystrike

Re: Logarithm

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

21. Dec 3, 2008

### Staff: Mentor

Re: Logarithm

BTW, if y isn't an integer, e^x doesn't mean x factors of e.

Assuming y > 0,
$$x = ln(y) \iff y = e^x$$

That's the relationship between these two functions. From this relationship, you can compose them in either order:
$$ln(e^x) = x$$
$$e^{ln(y)}= y$$ if y > 0.

22. Dec 3, 2008

### Sjorris

Re: Logarithm

Ok, if I describe both functions and then describe the total equation, you might figure it out.
ln(x)=a is a function which, when applied to x, returns a value a which. This value a has the property that if you powered e to to that value a would return x. So if ln(x)=a is true, then ea=x is true.

Now, in the equation you power e to a value ln(x), which we'll simplify to b for the moment, so ln(x)=b. Then the equation reads eb=x. We just established that ln(x)=b means that if you powered e to b, it would return x. And that's exactly what you are doing, you are powering e to b, or e to ln(x), so the outcome is x.

PS. This is true for x>0, otherwise you are dealing with complex numbers.

23. Dec 4, 2008

### HallsofIvy

Re: Logarithm

No, that's not true. I recommend you look up the definitions of ex and ln(x). There is no point in trying to do problems if you don't know the definitions of the very things you are working with!