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Logarithim Problem I been stuck on for a day

  • Thread starter lionely
  • Start date
576
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1. Homework Statement
I have been trying to solve this problem all day, I have used up 3 pages and to no avail.
I don't think i should try and change the bases because of the way the book is set up changing bases is not until the next chapter. So I want to try and get it out without changing bases. Can someone give me a hint?


2433yp2.jpg


The question is number 8
2. Homework Equations


3. The Attempt at a Solution
My work is 3 pages of scratch work so it is hard to type out but all I have done is try to combine the logs shown into 2 separate logs for the base a and the base 10. I get stuck there since I can't remove the logs.
 

pasmith

Homework Helper
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If the left hand side is [itex]\log_{a}(A) - \log_{10}(B)[/itex], what values do you have for [itex]A[/itex] and [itex]B[/itex]?
 
576
2
I got it down to A being (6/25) and B being 24
 

pasmith

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I got it down to A being (6/25) and B being 24
You have simplified too far; consider [tex]\log_a (6/25) - \log_{10}(24) = \log_a(24/100) - \log_{10}(24) = \log_a(24) - 2\log_a(10) - \log_{10}(24).[/tex]
 
576
2
I'm still lost because the bases are different.
 
576
2
-2 is equal to log(0.01) [ I'm using Log for the base a , because I don't know Latex]

so Log(24) - 2Log(10) = log(0.01) + log 24
 
19,164
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-2 is equal to log(0.01) [ I'm using Log for the base a , because I don't know Latex]

so Log(24) - 2Log(10) = log(0.01) + log 24
Now apply the product rule to log(0.01) + log (24). What do you get?
 
576
2
so Log(24) - 2Log(10) = log(0.24)
 
576
2
Log(24)- 2Log(10) = log(6/25)
 
576
2
Oh I see it was just something simple as that . I was thinking I would have to remove the log in the end never thought that I would just have to compare the bases....
So it was really easy I was overthinking it.
Thank you so much .

a is 10.
 

Ray Vickson

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I'm still lost because the bases are different.
Convert everything to ##\log_{10}##. There are standard formulas for converting logarithms between different bases.
 

SammyS

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Convert everything to ##\log_{10}##. There are standard formulas for converting logarithms between different bases.
OP said he was avoiding using that.
 

SammyS

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Problem (8) is equivalent to solving ##\ \log_a(A)=\log_{10}(A) \ ## for a.

Suppose it had been equivalent to solve ##\ \log_a(A)=\log_{10}(A) \ ## for a, where A ≠ B . How would you do that without the change of base formula?
 

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