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Base b of n exponent

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Updated 2007-11-12 00:45

Base 2 of n exponent

You may notice a pattern developing in the higher orders. For the higher powers it's more important to notice the millionth and billionth place as opposed to memorizing all the digits
QuestionAnswer
201
212
224
238
2416
2532
2664
27128
28256
29512
2101,024
2112,048
2124,096
2138,192
21416,388
2201,048,576
2301,073,741,824
2401,099,511,627,776
memorize

Base b (>2) of n exponent

 

These are common powers you may be expected to perform in your head.

 

Question Answer
329
3327
3481
4216
4364
5225
53125
6236
63216
7249
73343
8264
83512
9281
93729
102100
1031,000
1061,000,000
1091,000,000,000
10121,000,000,000,000
memorize

Logorithms

A logarithm of a given number to a given base is the power to which you need to raise the base in order to get the number.

 

In general n = log(x) (given the log's base equals b) because x = b^n
and for natural logs, n=ln(x) because x = e^n (the l in ln is lowercase L)
We generally assume that if the base isn't specified, the base is 10 (log(100) = 2 because 10^2 = 100)

 

Question Answer
log1002
log1,000,0006
log(base2)(1,000,000)~20 (or 19.931...)
e2.71828...
lnlog with base e
ln(e)1
ln(e^2)2
ln(e^3)3
e^(ln(b))b
e^(ln(4))4
e^(p*ln(b))b^p
e^(2*ln(3))3^2
ln(ab)ln(a) + ln(b)
ln(2*3)ln(2) + ln(3)
ln(c/d)ln(c) - ln(d)
ln(4/5)ln(4) - ln(5)
ln(55) = 4.007333...e^4.007333... ~= 55
(log(10))/(log(e)) = 1/(log(e))ln(10)
(ln(e))/(ln(10)) = 1/(ln(10))log(e)
memorize

Complex and negative exponents

 

Question Answer
e^(-a)1/(e^a)
e^(-2)1/(e^2)
e^(1/d)d'th root of e
e^(1/2)square root of e
e^(1/3)third (or cube) root of e
e^(2/5)fifth root of (e^2)
e^(6/8)eighth root of (e^6); or fourth root of (e^3)
e^(-2/3)1/(third root of (e^2))
8^(-2/3)1/(third root of (8^2)) = 1/4 = 1/(3rd rt of 64) = 1/((3rd rt of 8)^2)
9^(3/2)(square root of 9)^3 = 27 = 3^3 = square root of (9^3)
(e^a)*(e^b)e^(a+b)
(e^c)/(e^d)e^(c-d)
[(e^2)*(e^3)]/(e^4)e^((2+3)-4) = e
(e^2)/[(e^3)*(e^4)]e^(2-(3+4)) = e^(-5) = 1/(e^5)
memorize

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