Logarithim Problem I been stuck on for a day

[lionely]

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

Homework Equations

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]

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]

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]

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]

-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]

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]

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]

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]

Convert everything to $\log_{10}$. There are standard formulas for converting logarithms between different bases.

OP said he was avoiding using that.

 

[Ray Vickson]

OP said he was avoiding using that.

Right: I missed that.

 

[SammyS]

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?