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.

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)] 
--------------------------- 

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)