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I am sure this is a really dumb question but I am having trouble understanding it since I do not have any math background.

I have the logarithms of 2 values:

log(a) = 1347
log(b) = 1351

I am trying to calculate this:

exp( log(a) ) - exp( log(0.1) + log(b) )

I guess by transforming this into a sum/difference of logarithms and working in log scale because when I exponentiate the values become infinite. Basically I am looking for a solution for this in log scale, so that I do not get an Infinite value when entered in a calculator.

Does this make sense?


Thank you!


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That is just $a / (0.1 \cdot b)$ by basic properties of logaritms. Or am I overlooking something? –  vonbrand Feb 3 '13 at 19:58
I do not have a because I cannot solve for it since exp(log(a)) = Infinite. That is why I was looking to transform everything in log form.. is this correct: log(a)-(0.1*log(b)) ? –  Francesca Feb 4 '13 at 0:54

2 Answers 2

Hint: Recall that $$\exp(\log(a))=a\quad\text{and}\quad\exp(x+y)=\exp(x)\exp(y).$$ Here we're using the convention that the base of $\log$ is $e$.

Update: Using the first hint, we have $$\exp(\log a)=e^{1347}.$$ Combining the the second term of the equation, we have $$ \frac{1}{10}\exp(\log b)=\frac{1}{10}\exp({1351}).$$Now, combining the expressions, $$\exp(\log a)-\exp(\log(1/10)+\log(b))=e^{1347}-\frac{1}{10}e^{1351}.$$

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Hi Clayton, thank you. I only have the log(a) though and if I try to exponentiate I get Inf… –  Francesca Feb 3 '13 at 14:01
@Francesca: Since we know $\log(a)=1347$, we can deduce that $$a=\exp(1347).$$ If you try to put this in a calculator, it surely will throw an error, but you can use this. Does clarify anything for you? –  Clayton Feb 3 '13 at 14:03
$@Francesca: I've updated the solution. –  Clayton Feb 3 '13 at 14:12
A yes, thank you very much! –  Francesca Feb 3 '13 at 14:13
If you like and understand the answer, feel free to accept it. If not, keep asking questions until you understand it clearly –  Clayton Feb 3 '13 at 14:15

We're assuming $\;\log\;$ denotes $\;\log = \ln\;$, i.e., $\log$ base $e$, where $\exp(x)$ and $\log_e(x)$ are inverse functions.

Then remember that


So $\exp( \log a + \log b) = \exp(\log a)\exp(\log b)= a\cdot b$

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Like it+$$$$$$$$ –  Babak S. Feb 3 '13 at 15:24
Hi amWhy, thank you for your answer, but I cannot compute a = exp(log(a))), since this is equal to Inf, that is why I was looking for a solution in log terms... –  Francesca Feb 3 '13 at 15:31

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