# basic math question: transform a sum of exponents to a sum of logarithms

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?

 log(a)-(0.1*log(b))


Thank you!

-fra

-
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

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, $$\exp(\log a)-\exp(\log(1/10)+\log(b))=e^{1347}-\frac{1}{10}e^{1351}.$$

-
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 $$\exp(\log(a))=a\;\;\text{and}\;\;\exp(b+c)=\exp(a)\exp(c).$$ So$\exp( \log a + \log b) = \exp(\log a)\exp(\log b)= a\cdot b\$

-
Like it+ – S. Snape 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
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8137403726577759, "perplexity": 533.530693842931}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-07/segments/1454701157075.54/warc/CC-MAIN-20160205193917-00280-ip-10-236-182-209.ec2.internal.warc.gz"}