Tazz is explaining logarithms to Noah. Shall I give you an example?
In a nut shell: What power do I have to raise 2 in order to get 64? Or, 2
x = 64, solve for x. In this question is the basis of a logarithms. The 2
x = 64 can also be expressed, x = log
2( 64 ) = 6, because 2
6 = 64.
Which is bigger: 2
64 or 10
40 Classic pre-calc problem, and the problem Tazz is explaining to Noah.
To solve, we will use the identity
ln( ax ) = x ln( a ) ln( 264 ) = 64 * ln( 2 )
ln( 1040 ) = 40 * ln( 10 )
ln( 2 ) ≈ 0.69
ln( 10 ) ≈ 2.3
64*0.69 = 44.16
40 * 2.3 = 92
44.16 < 92
Therefore, 2
64 < 10
40 Computing 2
64 isn't enough to overflow a calculator with 20 digits. But if you wanted to see if 3
1000 is larger then 4
700 then using a logarithm is a better option.
How many digits are in 3
1000? Again... let's use a logarithm. Log in base 10, or log
10, will tell us how many digits are in a number. Why? Think about this. 10
3 = 10 * 10 * 10 = 1000--4 digits. 10
8= 100,000,000--9 digits. So digits =
%u230Alog
10( n )
%u230B + 1.
10x = 31000
log( 10x ) = log( 31000 ) Use identity : log( ax) = x log( a )
x * log( 10 ) = 1000 * log( 3 ) log( 10 ) = 1
x = 1000 * log( 3 )
log( 3 ) ≈ 0.4771
x = ( 1000 * 0.4771) = 477.13
%u230Ax%u230B + 1 = 478
So 3
1000 has 478 digits. A more general formula is for the digits in a
x is d = %u230Aa * log
10( x )%u230B + 1
One last problem, which I've had to do: what is 3.14
2.71? .
x = 3.142.71
ln( x ) = ln(
3.14
2.71 )
Use identity : ln( ax) = x ln( a )
ln( x ) = 2.71 * ln( 3.14 )
x = e
2.71 * ln( 3.14 )ln( 3.14 ) = 1.442
x = e2.71 * 1.442 = 22.22
If you are clever, you may have asked "That's great, but how do you calculate ln and e?". Turns out that this isn't all that hard, but that's for an other article.
