How to use logarithm

  • Thread starter DB
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  • #1
DB
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my physics teacher told us that for now well be solving the equation of a line such as [tex]y=kx^n[/tex] by trial and error for finding "n" using the multiplicative change in 2 points i.e [tex]x_1,y_1[/tex],[tex]x_2,y_2[/tex].
what we would do is for example 2 points (13,1) (6, 1.4):
[tex]n=\frac{y_2}{y_1}=(\frac{x_2}{x_1})^n[/tex]
[tex]n=\frac{6}{13}=(\frac{1.4}{1})^n[/tex]
so:
[tex]~0.46=1.4^?[/tex]
we'd have to use trial and error...wat id like to know is how to use logarithm to solve this, he said he might teach it but i was wondering if u guys could fill me in on how to solve this using log:
[tex]~0.46=1.4^n[/tex],
the answer for n is about -2.3...
thanks in advance
 

Answers and Replies

  • #2
zwtipp05
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You need to know the properties of logarithms
One is log a^x=x log a

Try looking in your precalc or algebra II book in order to see all the properties.
 
  • #3
DB
501
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i have none of those books, I am in sec 5 (grade 11)
plus i don't understand how to use the property u gave me :frown:
 
Last edited:
  • #4
amcavoy
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Suppose [itex]a=e^b[/itex]. Then [itex]a^{x}=e^{bx}[/itex] which implies [itex]\ln{a^{x}}=\ln{e^{bx}}=bx[/itex]. But since b=ln(a), you can substitute which gives [itex]\ln{a^{x}}=x\ln{a}[/itex].

Try taking the logarithm of both sides of your equation and use this property with some simple algebra. It should work out.
 
  • #5
TD
Homework Helper
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Well the answer isn't exactly -2.3
If you want to use logarithms, it goes like this:

[tex]0.46 = 1.4^n \Leftrightarrow \log 0.46 = \log 1.4^n \Leftrightarrow \log 0.46 = n\log 1.4 \Leftrightarrow n = \frac{{\log 0.46}}
{{\log 1.4}} \approx - 2.307853977[/tex]

Since I don't know how much you know, just tell us which steps you don't understand to get further details. As said before, we use that [itex]\log a^x = x\log a[/itex].
 
  • #6
DB
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i still don't get it, maybe if u could use whole numbers as examples?
 
  • #7
TD
Homework Helper
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It goes exactly the same, I always left the numbers so that wouldn't matter. Only in the last step I made an aproximation but that was done with a computer, you can't do it on your own for arbitrary numbers. To have the exact answer, you just leave the logarithm form.
 
  • #8
DB
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thanks guys i got it now, this will save me a lot of guessing :)
 
  • #9
DB
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just a question, why are we aloud to bring the exponent "n" down in front of the log as in:
log(0.46)=log(1.4)^n = log(0.46)= n log(1.4)
and why then can "n" be brought to the left side of the equation without turning negative?
 
  • #10
hypermorphism
506
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DB said:
just a question, why are we aloud to bring the exponent "n" down in front of the log as in:
log(0.46)=log(1.4)^n = log(0.46)= n log(1.4)
This was explained by apmcavoy in this thread.
DB said:
and why then can "n" be brought to the left side of the equation without turning negative?
We're dividing both sides of the equation by log(1.4).
 
  • #11
DB
501
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okay thanks hypermorphism I am going to study apmcavoy's thread, just a question about it: is he referring to the constan "e" ~ 2.7183?
 
  • #12
hypermorphism
506
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DB said:
... "e" ~ 2.7183?
Yep, that's the one. First discovered as the base of the natural logarithm function.
 
  • #13
HallsofIvy
Science Advisor
Homework Helper
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DB said:
just a question, why are we aloud to bring the exponent "n" down in front of the log as in:
log(0.46)=log(1.4)^n = log(0.46)= n log(1.4)
and why then can "n" be brought to the left side of the equation without turning negative?

Oh, my God, Oh, my God, Oh, my God.

If you are working with logarithms you should not even be asking a question like that! I wish there were a way of going back in time and killing your algebra teachers! (Unless of course you simply didn't pay attention to them. In which case, I should simply kill you!)

We do not ever just "bring a number to the left side"! We do the opposite of what ever operation is done to the number we want to isolate.

IF we have, for example, x+ a= b, then, since a is added to x, we do the opposite: subtract a from both sides so that we get x+a-a= b-a or x= b- a. Yes, in this one case, the number "a" has "changed sign" or "turned negative".

IF we have, for example, ax= b, then, since x is multiplied by a, we do the opposite: divide by a. ax/a= b/a or x= b/a. There is no "negative" involved in that!

IF we have, for example, ex= b, then since x is an exponential, we do the opposite: take the logarithm so that log(ex)= log b or
x= log b. log(ex)= x because that is the way log is defined!
There is no "negative" involved!

Sorry if I sound over the top here but I am teaching two sections of "remedial algebra" and I spend most of my time cursing high school algebra teachers!
 
  • #14
DB
501
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hallsofivy, relax, i know my algebra, to be honest u've really insulted me by "teaching" me wat you did. i simply just thought that an "n" in front of a log (n log(1.4)) ment something other than multiplication until hypermorphism told me about dividing both sides and i saw the algebra automatically, realizing it was multiplication. I've never used a log before in my life neither the constant "e" so i didnt know wat to expect. I've got no reason to curse my algebra teachers, i do pretty good in it.
 
  • #15
DB
501
0
im currently in 536 math...do u really think i would be in it if i couldn't do the examples you posted? i sure as hell would have trouble with quadratics, proofs and optimization...
and i really don't care who you want to kill

by the way its 500 + 30 + 6, just in case you didnt know how those numbers came to be...
 

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