SOLUTION: Please help me with these questions ASAP!! Suppose that f(x)=log3(x-1) a)If f(6)=1.4 and f(8)=1.7,then evaluate log3(35) and log3(7/5). b)If f(3)=0.6, find log9(2).

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Please help me with these questions ASAP!! Suppose that f(x)=log3(x-1) a)If f(6)=1.4 and f(8)=1.7,then evaluate log3(35) and log3(7/5). b)If f(3)=0.6, find log9(2).       Log On


   



Question 276590: Please help me with these questions ASAP!!
Suppose that f(x)=log3(x-1)
a)If f(6)=1.4 and f(8)=1.7,then evaluate log3(35) and log3(7/5).
b)If f(3)=0.6, find log9(2).

Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
f%28x%29=log%283%2C+%28x-1%29%29
f%286%29=log%283%2C+%286-1%29%29+=+log%283%2C+%285%29%29
We're told that f(6) = 1.4 so log%283%2C+%285%29%29+=+1.4
f%288%29=log%283%2C+%288-1%29%29+=+log%283%2C+%287%29%29
We're told that f(8) = 1.4 so log%283%2C+%287%29%29+=+1.7
a) log%283%2C+%2835%29%29
Since the only base 3 logs we know are log%283%2C+%285%29%29 and log%283%2C+%287%29%29, we need to express log%283%2C+%2835%29%29 in terms of the other two. Since 35 = 7*5 we can start with
log%283%2C+%2835%29%29+=+log%283%2C+%287%2A5%29%29
This does not mean that log%283%2C+%2835%29%29+=+1.7%2A1.4. But we can use a property of logarithms, log%28a%2C+%28p%2Aq%29%29+=+log%28a%2C+%28p%29%29+%2B+log%28a%2C+%28q%29%29, to separate our log into two separate logs:

For log%283%2C+%287%2F5%29%29 we'll use another property of logarithms: log%28a%2C+%28p%2Fq%29%29+=+log%28a%2C+%28p%29%29+-+log%28a%2C+%28q%29%29
log%283%2C+%287%2F5%29%29+=+log%283%2C+%287%29%29+-+log%283%2C+%285%29%29+=+1.7+-+1.4+=+0.3

b) f%283%29+=+log%283%2C+%283-1%29%29+=+log%283%2C+%282%29%29
Since we're told that f(3) = 0.6, log%283%2C+%282%29%29+=+0.6.
For log%289%2C+%282%29%29 we do not know any base 9 logs and 2 is not an obvious power of 9. So what so we do? Well we do know several base 3 logarithms so we'll use the base conversion formula formula, log%28a%2C+%28p%29%29+=+log%28b%2C+%28p%29%29%2Flog%28b%2C+%28a%29%29, to change this base 9 logarithm into an expression of base 3 logarithms:
log%289%2C+%282%29%29+=+log%283%2C+%282%29%29%2Flog%283%2C+%289%29%29
From above we know the numerator and the denominator can be done "by hand" since 9 is a well known power of 3:
log%289%2C+%282%29%29+=+log%283%2C+%282%29%29%2Flog%283%2C+%289%29%29+=+0.6%2F2+=+0.3