Simple Logs

  • Thread starter cscott
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  • #1
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[tex]log_x 4 + log_x 8 =5[/tex]

I simplified that to [itex]log_x 32 = 5[/itex] but I can't get my head around what to do next and it's annoying me because I feel it's going to be something simple. :smile:
 

Answers and Replies

  • #2
Curious3141
Homework Helper
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cscott said:
[tex]log_x 4 + log_x 8 =5[/tex]

I simplified that to [itex]log_x 32 = 5[/itex] but I can't get my head around what to do next and it's annoying me because I feel it's going to be something simple. :smile:
What you have isn't that much of a simplification, it would be better to break everything on the LHS to powers of 2 and get smaller numbers. Nevertheless, you can solve the equation immediately from what you have.

What is the definition of a logarithm ? What does it mean when you say [tex]log_a b = c[/tex]. Come up with a simple equation to relate a, b and c using the definition of log, then apply the same principle to your equation, and see what you get.
 
  • #3
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hey there since the log is to the base x, it follows that
x^(LOGx32)=x^5 the x and the LOGx cancel out the we have
=>32=x^5
therefore x=2
 
  • #4
782
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Thanks, I understand that now, but I'm having trouble with another :mad:

[tex]20 000 = 10 000(x)^9[/tex]
[tex]2 = x^9[/tex]
[tex]log 2 = log x^9[/tex]
[tex]log 2 = 9 \cdot log x[/tex]

I think I got closer to the answer...
 
  • #5
782
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steven187 said:
hey there since the log is to the base x, it follows that
x^(LOGx32)=x^5 the x and the LOGx cancel out the we have
=>32=x^5
therefore x=2
Is there any way to do it without using trial an error in the end (if the numbers weren't as nice as they are here)?
 
  • #6
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cscott said:
Thanks, I understand that now, but I'm having trouble with another :mad:

[tex]20 000 = 10 000(x)^9[/tex]
[tex]2 = x^9[/tex]
[tex]log 2 = log x^9[/tex]
[tex]log 2 = 9 \cdot log x[/tex]

I think I got closer to the answer...
Actually, once you've got [tex]2 = x^9[/tex], all you've got to do is [tex]\sqrt[9]{2} = x = 1.080059739[/tex].

cscott said:
Is there any way to do it without using trial an error in the end (if the numbers weren't as nice as they are here)?
I don't think it's really trial and error; had it been 31 or 33, the answer wouldn't have been an integer. "Luckily", [tex]\sqrt[5]{32} = 2[/tex].
 
  • #7
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They design textbook problems so you dont have to go through that mess, just understand the cocnept and know how to repeat the steps you just went through and you'll be fine.
 
  • #8
James R
Science Advisor
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Just remember, that

[tex]\log_a b[/tex]

is the power which you must raise a to in order to get b. So

[tex]\log_2 32 = 5[/tex]

Equivalently, if

[tex]\log_a b = x[/tex]

then

[tex]b = a^x[/tex]
 

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