Can anyone please help me with these log problems?

Question

http://online.math.uh.edu/Math1310/ch5/s54/LawLog/List/images/List5.gif (you dont have to do the calculator part)

and

http://online.math.uh.edu/Math1310/ch5/s54/LawLog/List/images/List6.gif

THANKS. Please show the steps?

Answer 1

1)
log(base6)(40)= log(base10)(40)/log(base10)(6)

2) Factor 42 to 7*3*2.

ln(42)=ln(7*3*2)=ln7+ln3 + ln2

Factor 45 to 3*3*5 .

log45=log(3*3*5)=2log3+log5

Answer 2

The first is a simple application of the change of base formula.

log base 6 of 40 = log base 10 of 40 / log base 10 of 6.

The second one just asks you to show that you understand how multiplication and addition of logs correspond to exponentiation and multiplication.

So

log 45 = log 9(5) = log 9 + log 5 = log 3² + log 5 = 2log 3 + log 5.

You do the other one.

Answer 3

1) Log base a of b = log b / log a
log base 6 of 40 = log 40 / log 6

2a) 42=3*2*7 and log ab = log a + log b, so ln 42 equals ln 2 + ln 3 + ln 7

2b) Similarly, 45=3*3*5, and log 45 equals log 3 + log 3 + log 5

Answer 4

log base 6 of 40 = (log_6)(40)= ln(40)/ln(6)
This follows directly from the change of basis formula.
ln(42) = ln(2) + ln (3) + ln (7)
ln(45) = 2*ln(3) + ln(5)
This follows from the fact that the logarithm of a product of primes is equal to the sum of the logarithm of each prime.