Logarithm Problem
- Thread starter Amelia Hu
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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 \(H\)?
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 \(H\)?
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.
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.
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
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If \(\displaystyle log_{10}A = log_{10}B\) then \(\displaystyle A=B\). Does that help?
HallsofIvy
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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.
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.
\(\displaystyle log_{10} \left ( \dfrac{x}{y} \right ) = 4 = 4 * 1 = 4 * log_{10}(10) = log_{10}(10^4) \implies\)
\(\displaystyle \dfrac{x}{y} = 10^4 = 10,000.\)
The above takes advantage of three laws of logs.
\(\displaystyle log_a(a) = 1.\)
\(\displaystyle b * log_a(c) = log_a(c^b).\)
\(\displaystyle log_a(x) = log_a(y) \implies x = y.\)
\(\displaystyle \dfrac{x}{y} = 10^4 = 10,000.\)
The above takes advantage of three laws of logs.
\(\displaystyle log_a(a) = 1.\)
\(\displaystyle b * log_a(c) = log_a(c^b).\)
\(\displaystyle log_a(x) = log_a(y) \implies x = y.\)
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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