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I am learning for a math exam and have the following solution:

$$ 0.01 = 0.5^n\\ n \cdot \log 0.5 = \log 0.01\\ n=\frac{\log 0.01}{\log 0.5} $$

OK, so far, so good. (I guess)

But now, it gets weird:

$$ n=\frac{\log 0.01}{\log 0.5}=\frac{0-2}{0.7-1}=… $$

Can somebody please explain how to go from $\log 0.01$ to $0-2$ and from $\log 0.5$ to $0.7-1$?

Daniel Hilgarth
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6 Answers6

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log0.01 = log 1/100 and log 0.5 = log(5/10) now apply log(m/n) = log m -log n

john
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log 0.01 = log(1/100) = log 1 - log 100 = log 1 - 2 log 10 = 0-2

log 0.5 = log(2/10) = log 2 - log 10 = 0.7 - 1

log(a/b) = log a -log b

log(a^n) = n log a

Hope this helps!

Kush Shukla
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$$n=\frac{log0.01}{log0.5}$$.Remebner that $logm^n=mlogn$.Therefore $log 0.01=log$ $\mathrm{10}^{-2}$=$-2$.$log.5=log \frac{1}{2}=log1-log2=0-log2$

Devgeet Patel
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$$n \ = \ \frac{log_{10} \frac{1}{100}}{ log_{10} \frac{5}{10}} $$

Can you finish it now?

neofoxmulder
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For the numerator, notice that $.01 = \frac{1}{100}$ and by logarithmic rules $$\log .01=\log\frac{1}{100}=\log1-\log100=0-2$$ The same goes for the denominator: $.5=\frac{5}{10}$ $$\log .5=\log \frac{5}{10}=\log5-\log10=0.7-1$$

Using the logarithmic rule: $$\log_a\left(\frac{m}{n}\right)=\log_a m-\log_an$$

Zhoe
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  • Nice, thanks. How would I know that log 5 = 0.7? The 0.7 is not even exact, right? – Daniel Hilgarth Feb 15 '14 at 14:23
  • Are you allowed to use calculators? You'd get $\log 5=0.69897...\approx .7$. – Zhoe Feb 15 '14 at 14:24
  • Yeah, I am allowed. But with a calculator, I could just punch in $\log 0.5$ directly... – Daniel Hilgarth Feb 15 '14 at 14:25
  • I would have done the same..I am not sure why the solution bothers to expand, but you'll get the same answer. – Zhoe Feb 15 '14 at 14:27
  • OK, thanks! :-) – Daniel Hilgarth Feb 15 '14 at 14:57
  • If you have a $\sqrt{}$ button but no $log$ button you can do this ,

    $ \sqrt{10} \approx 3.16 $ , $ \sqrt{ \sqrt{10} } \approx 1.78 , $ \sqrt{ \sqrt{ \sqrt{ \sqrt{10} }}} \approx 1.15$

    $$ \frac{(3.16)(1.78)}{1.15} \approx 5$$

    $$ \frac{10^{\frac{1}{2}} \cdot 10^{ \frac{1}{4}}}{10^{ \frac{ 1}{16}}} \approx 10^{ \frac{11}{16}} $$

    $$\frac{11}{16} \approx 0.7$$

    So we get close to the exponent in $10^x = 5$ , $x \approx 0.7$

     – neofoxmulder Feb 15 '14 at 15:04
  • Ah , i ran out of time to edit. $\frac{(3.16)(1.78)}{1.15} \ \approx \ 5 $ . Can you figure out the rest? :) – neofoxmulder Feb 15 '14 at 15:18
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As mentioned above, it comes from the properties of quotients of logs. It looks like the way logs were written when people looked them up in tables (the tables went from $\log 1$ to $\log 9.999$, and you added/subtracted the exponent (as if you had written the number in scientific notation)), but I have no idea if they're actually expecting you to use tables here.

tabstop
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