# Solve 10 base logarithms

I'm a n00b in math and I wanted to know how should I solve ten base logarithms. e. g.:

Log 40 to base 10

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Take a pocket calculator, type '40' and press 'log'. –  Fabian Jan 11 '12 at 18:44
@MahanGM. What kind of solution do you want? Cant you use the calculator? –  smanoos Jan 11 '12 at 18:45
I actually did but I didn't know that the calculators solve in 10 base automatically. Thanks for reply. –  user22780 Jan 11 '12 at 18:46
@MahanGM. They do. Check Fabian's comment. –  smanoos Jan 11 '12 at 18:49
FWIW, we "solve" equations, inequalities, problems etc. We don't "solve" logarithms, we evaluate them, manipulate them etc. –  Daniel Freedman Jan 11 '12 at 18:56

1.602059991327962390427477789448986053536379762924217082620854922254216378548849018973854504236372344081368954382861...

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To get a very close approximation, you're going to need a calculator. A not-so-accurate one gives 1.602059991.

For a quick order of magnitude calculation, it suffices to remember sqrt(10) ~ 3 adds 0.5 to logs in base 10, sqrt(3) ~ 1.7 adds 0.25 to logs in base 10, and sqrt(1.7) ~ 1.3 adds 0.125 to logs in base 10, and possibly so on. So in this case, 40 ~ 10 * 3 * 1.3, so log(40) = log(10*3*1.3) = 1+0.5+0.125 = 1.625, not a bad estimate when you can get it off-hand.

If you want something to approximate the answer on a napkin, you can do a Taylor series expansion to a few terms. d(log_10 x)/dx = d(ln x / ln 10)/dx = 1/(ln(10) x). At = 10, we have f(x) = 1 and f'(x) = 1/10ln(10). So all we need to compute is 1 + (40 - 10)/10ln10. 10 is a little more than e^2, so we can do 1 + 30/20 = 2.5. This approximation wasn't very good since it's first-order and 10 isn't particularly close to 40 (for the log_10 function, that is). If you had log_10(110), we would have gotten f(x) = 2 and f'(x) = 1/100ln(10), and found 2 + (110-100)/200 = 2+0.05 = 2.05... not to bad, considering the real answer is around to 2.041392685.

Naturally these approximation methods work well when, in the first case, you can decompose the number as a product of 10, 3, 1.7 and 1.3, and in the second case when the number you want to do is close to a number for which it's easy to evaluate the function and its derivative. Other approximation methods may exist.

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See this, Pat: codecogs.com/latex/eqneditor.php –  The Chaz 2.0 Jan 11 '12 at 19:09
Hey thanks, The Chaz! That will be handy. –  Patrick87 Jan 11 '12 at 19:12

$\log (40) = \log (4) + \log (10) = \log (4) + 1 = 2 \log (2) + 1$

So if you are given (as you might be on a test) a few logarithms (say log (2) ~ .3 and log (3) ~ .48), then you can use some manipulations to arrive at an approximation.

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