# Logarithim Problem I been stuck on for a day

#### lionely

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?

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.

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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 $B$?

#### lionely

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 $$\log_a (6/25) - \log_{10}(24) = \log_a(24/100) - \log_{10}(24) = \log_a(24) - 2\log_a(10) - \log_{10}(24).$$

#### lionely

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

#### lionely

-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?

#### lionely

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

#### lionely

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

#### lionely

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.

Homework Helper
Dearly Missed

#### SammyS

Staff Emeritus
Homework Helper
Gold Member
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?

"Logarithim Problem I been stuck on for a day"

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