# Logarithm Problem

#### Amelia Hu

##### New member
how to solve log 10 to the power of log (A-)/(HA) = log10 to the power of 4? Thanks

#### MarkFL

##### Super Moderator
Staff member
Hello, and welcome to FMH!

I am unclear what "A-" means...can you post a pic of the original equation, or format it more clearly? For example, I am unsure whether the exponent on the left side is:

$$\displaystyle \frac{\log(A-)}{HA}$$

or:

$$\displaystyle \log\left(\frac{A-}{HA}\right)$$

Also, are we to solve for $$A$$ or $$H$$?

#### Amelia Hu

##### New member
Hello, Thanks!

It's $$\displaystyle \log\left(\frac{A-}{HA}\right)$$ . So are to solve (A-)/(HA) together ? I thought of putting log10 on both sides. I do not know if log10 to the power of $$\displaystyle \log\left(\frac{A-}{HA}\right)$$, does the log10 and the log cancel out to be just (A-)/(HA) and the other side log10 to the power of 4 will be converted to the exponential which is 10 to the power of 4? Is that so that there is a rule of log10 and the log cancel out to be just (A-)/(HA)? Then (A-)/(HA) is (10 to the power of 4 ) over 1. Thanks. Do hear from you soon.

#### MarkFL

##### Super Moderator
Staff member
I still don't know what "A-" means.

#### Amelia Hu

##### New member
Hi, thanks. It's actually a chemistry question, so it is the ratio of A− (ionized) to HA (unionized) and I'm supposed to find the the ratio of A− (ionized) to HA (unionized). I'm just thinking of the mathematics part of log10 to the power of $$\displaystyle \log\left(\frac{A^-}{H_A}\right)$$ , does the log10 and the log cancel out to be just (A-)/(HA) ?

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#### Harry_the_cat

##### Senior Member
how to solve log 10 to the power of log (A-)/(HA) = log10 to the power of 4? Thanks
If $$\displaystyle log_{10}A = log_{10}B$$ then $$\displaystyle A=B$$. Does that help?

#### HallsofIvy

##### Elite Member
Okay, so "H-" and "HA" are just numbers that we could as well call "x" and "y" (for the non-chemists among us). And you are asking about $$\displaystyle log\left(\frac{x}{y}\right)$$?
But I am still not clear on what you mean by "log 10 to the power of". Do "logarithm to base 10" of "logarithm of 10"?

If you just mean "$$\displaystyle log\left(\frac{H-}{HA}\right)= -4$$" (the logarithm of $$\displaystyle 10^{-4}$$ is just -4) then use the fact that $$\displaystyle log\left(\frac{H-}{HA}\right)= log(H-)- log(HA)= -4$$ so that log(H-)= log(HA)- 4 or, equivalently, that log(HA)= log(H-)+ 4.

#### Amelia Hu

##### New member
Hi, Thanks. But I'm trying to ask what's in the picture :

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#### JeffM

##### Elite Member
$$\displaystyle log_{10} \left ( \dfrac{x}{y} \right ) = 4 = 4 * 1 = 4 * log_{10}(10) = log_{10}(10^4) \implies$$

$$\displaystyle \dfrac{x}{y} = 10^4 = 10,000.$$

The above takes advantage of three laws of logs.

$$\displaystyle log_a(a) = 1.$$

$$\displaystyle b * log_a(c) = log_a(c^b).$$

$$\displaystyle log_a(x) = log_a(y) \implies x = y.$$

#### Amelia Hu

##### New member
Hi, thanks a lot. But then to be exact, it is not log 10 ( x / y ), it is more of log ( x/y). Does it in this does not make sense?

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#### Amelia Hu

##### New member
Hi, thanks a lot. But then to be exact, it is not log 10 ( x / y ), it is more of log ( x/y). Does it in this does not make sense?
Hi, thanks a lot. You're right. The text has some printing error , so it's log10 (x/y). It's solved

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