Properties of Logs: What is the Relationship Between ln(x) and ex?

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Homework Help Overview

The discussion revolves around the properties of logarithms and their relationship with exponential functions, specifically focusing on the natural logarithm and the base \( e \). Participants are attempting to demonstrate a specific mathematical relationship involving logarithms and exponents.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss various properties of logarithms, including the product rule and the power rule. There are attempts to apply these properties to the problem at hand, with some questioning the validity of applying logarithmic rules to certain expressions.

Discussion Status

There is an ongoing exploration of logarithmic properties, with some participants providing guidance on how to apply these rules. Multiple interpretations of the problem are being considered, and there is a mix of understanding and confusion regarding the relationship between logarithmic and exponential functions.

Contextual Notes

Some participants express uncertainty about the definitions and properties of logarithms and exponentials, indicating a need for clarification on these foundational concepts.

icystrike
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Homework Statement


the question is this:
showhttp://img80.imageshack.us/img80/8628/qnszb4.jpg
is 2[tex]\sqrt{3}[/tex]

whereby e is natural logarithm base

Homework Equations


The Attempt at a Solution

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


What do you know about the properties of logs?
 


Integral said:
What do you know about the properties of logs?

i suppose everything.. i just can't show tat it is 2sqrt3
 


Do you have a text?

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


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

i think tats all of it?
 
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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.
 


Integral said:
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.

but it is just the exponent constant we can't apply the logarithm rules unless to the power.
 


Repeat in english please
 
  • #10


oh.. let me rephrase
the equation is just the the sums of 2 exponent constant thus we can't apply the any logarithm rules except to the power of the constant
 
  • #11


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.


[tex]e^{xlny} = e^{lny^x}[/tex]
 
  • #12


Integral said:
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.[tex]e^{xlny} = e^{lny^x}[/tex]

yup i applied already
3e^(ln (1/sqrt3) ) + e^(ln sqrt3)
 
  • #13


icystrike said:

Homework Statement


the question is this:
showhttp://img80.imageshack.us/img80/8628/qnszb4.jpg
is 2[tex]\sqrt{3}[/tex]

whereby e is natural logarithm base

Homework Equations





The Attempt at a Solution


You need: a ln(b)= ln(ba) and eln(x)= x.
 
Last edited by a moderator:
  • #14


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

now you are getting some where.

you do know that :

[tex]e^{lnx} = x[/tex]
 
  • #15


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

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

thank you HALLSOFIVY and INTEGRAL .:smile:
mind explaining this law eln(x)= x ?
 
  • #16


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

then understand that ln = loge
 
  • #17


Integral said:
On the page YOU linked read the POWER RULE of logs.

then understand that ln = loge

ya.. i know about it.
how about this?
e^ln(x)= x
 
  • #18


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


Sjorris said:
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.

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


icystrike said:
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?

btw you see, ln x means x=e^? and e^x means e x e x e x e x ...
^
l
x factors
i don't really see the relationship
 
  • #21


icystrike said:
btw you see, ln x means x=e^? and e^x means e x e x e x e x ...
^
l
x factors
i don't really see the relationship
BTW, if y isn't an integer, e^x doesn't mean x factors of e.

Assuming y > 0,
[tex]x = ln(y) \iff y = e^x[/tex]

That's the relationship between these two functions. From this relationship, you can compose them in either order:
[tex]ln(e^x) = x[/tex]
[tex]e^{ln(y)}= y[/tex] if y > 0.
 
  • #22


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


icystrike said:
btw you see, ln x means x=e^? and e^x means e x e x e x e x ...
^
l
x factors
i don't really see the relationship
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!
 

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