Help with Precalculus honors math problem?

Question

Please help I've done this problem tons of times and I can't figure it out!!! Thanks!

Approximate the logarithm using the properties of logarithms and given that
log(base b)2 = .3562
log(base b)3 = .5646
log(base b)5 = .8271

70. log(base b) (25/9)

70 isnt part of the problem it's the number question sorry!

Answer 1

The trick here is the use the exponential and division properties for logarithms:

Exponents: a log b = log b^a.
Division: log a/b = log a - log b

Step 1: Convert the fraction to 2 logarithms.
log (b) 25/9 = log (b) 25 - log (b) 9

Step 2: Convert each logarithm to exponents, and then pull the exponent out.
log (b) 25 = log (b) 5^2 = 2 log (b) 5
log (b) 9 = log (b) 3^2 = 2 log (b) 3

Step 3: Use the given values of log (b) 3 and log (b) 5.
log (b) 25/9 = 2 log (b) 5 - 2 log (b) 3
log (b) 25/9 = 2 (.8271) - 2 (.5646)
log (b) 25/9 = 1.6542 - 1.1292
log (b) 25/9 = .5250 (solution!)

Answer 2

25 is 5 squared, and 9 is 3 squared
so, you can change the problem to
70log(baseb)(5^2/3^2)
then
70(2)log(baseb)5 - 70(2)log(baseb)3
140((log baseb)5- log(baseb)3)
140(.8271- .5646)
simplify

Answer 3

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