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You Can Do Math: Scientific Notation, Engineering Notation and Standard Form?
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In chapter 7 of my book on scientific notation, I talk about how to approx calc logs of numbers in your head.
Using this technique
Log(0.95) = log( 9.5 * 10^-1 ) = 0.98(approx) - 1 = -0.02 (approx)
Assuming you know how to use logs (i havent written a book on logs yet!), and then use the technique backwards, you can get an answer.
Log( 0.95^10 ) = 10 * log(0.95) = 10 * -0.02 = -0.2
Now -0.2 = -1 +0.8
Using the technique backwards:
10^-0.2 = 6.5 * 10^-1 = 0.65 (approx)
Actual answer is 0.599
Why i am off?
Mainly: log(0.95) is actually -0.022 my first guess interpolation between log(0.9) and log(1) was a little off. So 10*log(0.95) = -0.22 = -1 + 0.78. And using the technique backwards would give 10^-1+0.78 = 6 * 10^-1 = 0.6
Book link:
As long as the base (the thing being raised to a power) is the same
when multiplying add the exponents so a^5 * a^3 = a^8 and 10^5 * 10^3 = 10^8
when dividing subtract exponents so a^5 / a^3 = a^2 amd 10^5 / 10^3 = 10^2
just do the other bits in the normal way - remember order of multiplication doesnt matter - and you can rearrange divisions
So 3 * 10^5 * 2 * 10^7 = ( 3 * 2 ) * ( 10^5 * 10^7 ) = 6 * 10^12
And ( 8 * 10^5 ) / ( 2 * 10^7 ) = ( 8 / 2 ) * ( 10^5 / 10^7 ) = 4 * 10^-2
You may find my kindle book useful
http://www.suniltanna.com/indices.php
I've also done one on scientific notation (also known as standard form) and engineering notation, which isnt on my website yet but is on amazon (search my name "Sunil Tanna") or
Book link:
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