Question 276590
{{{f(x)=log(3, (x-1))}}}
{{{f(6)=log(3, (6-1)) = log(3, (5))}}}
We're told that f(6) = 1.4 so {{{log(3, (5)) = 1.4}}}
{{{f(8)=log(3, (8-1)) = log(3, (7))}}}
We're told that f(8) = 1.4 so {{{log(3, (7)) = 1.7}}}
a) {{{log(3, (35))}}}
Since the only base 3 logs we know are {{{log(3, (5))}}} and {{{log(3, (7))}}}, we need to express {{{log(3, (35))}}} in terms of the other two. Since 35 = 7*5 we can start with
{{{log(3, (35)) = log(3, (7*5))}}}
This does <b>not</b> mean that {{{log(3, (35)) = 1.7*1.4}}}. But we can use a property of logarithms, {{{log(a, (p*q)) = log(a, (p)) + log(a, (q))}}}, to separate our log into two separate logs:
{{{log(3, (35)) = log(3, (7*5)) = log(3, (7)) + log(3, (5)) = 1.7 + 1.4 = 3.1}}}
For {{{log(3, (7/5))}}} we'll use another property of logarithms: {{{log(a, (p/q)) = log(a, (p)) - log(a, (q))}}}
{{{log(3, (7/5)) = log(3, (7)) - log(3, (5)) = 1.7 - 1.4 = 0.3}}}<br>
b) {{{f(3) = log(3, (3-1)) = log(3, (2))}}}
Since we're told that f(3) = 0.6, {{{log(3, (2)) = 0.6}}}.
For {{{log(9, (2))}}} 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(a, (p)) = log(b, (p))/log(b, (a))}}}, to change this base 9 logarithm into an expression of base 3 logarithms:
{{{log(9, (2)) = log(3, (2))/log(3, (9))}}}
From above we know the numerator and the denominator can be done "by hand" since 9 is a well known power of 3:
{{{log(9, (2)) = log(3, (2))/log(3, (9)) = 0.6/2 = 0.3}}}