Question 29076: What is [log, base 3b(3b again but regular number)exponent 5] times [log, base 11 1]?
Solve for b: 7log b=3
Solve for d: 4ln(4d-2)=6
Please give me the steps so I know how to do them.
Answer by sdmmadam@yahoo.com(530)   (Show Source):
You can put this solution on YOUR website! What is [log, base 3b(3b again but regular number)exponent 5] times [log, base 11 1]?
Solve for b: 7log b=3
Solve for d: 4ln(4d-2)=6
Please give me the steps so I know how to do them.
log, base 3b(3b again but regular number)exponent 5] times [log, base 11 1]?
log[(3b)^5] X log(1) (using log(b)^m to the base (b) = m log(b) to the base b
(3b) 11
and log anything to the same base is 1 and hence 5X1 = 5 and log(1) to any base is zero )
=5 X 0
=0
Solve for b: 7log b=3
Quick way of doing this is dividing by 7 and then applying definition of logarithm of a positive number
7log b=3
Dividing by 7,we get
log(b)=(3/7)
(here the base 10 is understood )
b= =[(10)^(3/7)]
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Note:Students tend to do the same as follows which is round about.
7log b=3
log[(b)^7] = 3 (here the base 10 is understood )
[using definition log(N) to base b is the power p to which the base has to be raised to give the number. that is N=(b)^ ]
[(b)^7] =[(10)^3] [using definition log(N) to base b is the power p to which the base has to be raised to give the number that is N=(b)^p and here N should be a strictly positive number]
Taking the seventh root on both the sides.
[(b)^7]^(1/7) =[(10)^3]^(1/7)
b= =[(10)^(3/7)] [using (a^m)^n = (a)^(mn)]
3) 4ln(4d-2)=6
Dividing by 4
log(4d-2) = 6/4 =3/2 (cancelling 2 in the nr and in the dr)
log(4d-2)=(3/2)
implies (4d-2) = [(10)^(3/2)]
4d = [(10)^(3/2)]+2
Dividing by 4
d = (1/4)[(10)^(3/2)]+(1/2)
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