base^log base ( x )= x
log100 (2) = log 10(2) * log100(10)
but
100^log 100(10) = 10
so log100 (10) = .5
because
100^.5 = 10
so
log 100(2) = .30103*.5
=0.151 <------ answer
check
100^.151 = ?
2 sure enough
others similar - will continue but have to go now
1. To five decimals it is a fact that log(2)=.30103 Compute the following to three decimal places usuing ONLY this fact and the logarithmic rules and the definition of a logarithm.
a. log100 (2) --base 100
b. log1000 (square root of 2) -- base 1000
2. Suppose you know that logb(C) = D. If logb(E)=D^2, find E in terms of some (or all) of the terms b, C, and D.
2 answers
b.
log 1000 (2^.5) = .5 log 1000 (2)
log 1000(2) = log 10(2) log 1000(10)
log 1000(10) = 1/3
because 1000^(1/3) = 10
so
.5 log1000(2) = .5(.30103)/3
=.0502 <----- answer
check
1000^.0502 =?
1.414 sure enough sqrt 2
2. Well, it does not matter what base so I will use ten and just write log
D = log C
D^2 = log C * log C
but D^2 = log E
so
log E = log C * log C
so
10^log E = E = 10^(log C*log C)
so
E = [10^log C]^log C
but 10^log C = C
so
E = C^log C
but log C = D
so
E = C^D
remarkable :)
log 1000 (2^.5) = .5 log 1000 (2)
log 1000(2) = log 10(2) log 1000(10)
log 1000(10) = 1/3
because 1000^(1/3) = 10
so
.5 log1000(2) = .5(.30103)/3
=.0502 <----- answer
check
1000^.0502 =?
1.414 sure enough sqrt 2
2. Well, it does not matter what base so I will use ten and just write log
D = log C
D^2 = log C * log C
but D^2 = log E
so
log E = log C * log C
so
10^log E = E = 10^(log C*log C)
so
E = [10^log C]^log C
but 10^log C = C
so
E = C^log C
but log C = D
so
E = C^D
remarkable :)