# Logarithim Problem I been stuck on for a day

## 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?

The question is number 8

## 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.

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pasmith
Homework Helper
If the left hand side is $\log_{a}(A) - \log_{10}(B)$, what values do you have for $A$ and $B$?

I got it down to A being (6/25) and B being 24

pasmith
Homework Helper
I got it down to A being (6/25) and B being 24
You have simplified too far; consider $$\log_a (6/25) - \log_{10}(24) = \log_a(24/100) - \log_{10}(24) = \log_a(24) - 2\log_a(10) - \log_{10}(24).$$

I'm still lost because the bases are different.

Chestermiller
Mentor
I got it down to A being (6/25) and B being 24
You have $$\log_{a}(6/25)=\log_{10}24-2$$
Lets consider the -2 in the equation. What is ##\log_{10}0.01## equal to?

Chet

-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

Chestermiller
Mentor
-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?

so Log(24) - 2Log(10) = log(0.24)

Chestermiller
Mentor
so Log(24) - 2Log(10) = log(0.24)
Now express 0.24 as a fraction reduced to lowest terms.

Chet

Log(24)- 2Log(10) = log(6/25)

Chestermiller
Mentor
Log(24)- 2Log(10) = log(6/25)
So, now you have $$\log_{a}(6/25)=\log_{10}(6/25)$$
So, what is a equal to?

Chet

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
Homework Helper
Dearly Missed
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
Staff Emeritus
Homework Helper
Gold Member
Convert everything to ##\log_{10}##. There are standard formulas for converting logarithms between different bases.
OP said he was avoiding using that.

Ray Vickson
Homework Helper
Dearly Missed
OP said he was avoiding using that.
Right: I missed that.

SammyS
Staff Emeritus