1. ## Evaluating Logs

Hey guys, need help with this

1. Suppose that f(x) = $\displaystyle \log_{3} (x-1)$

a. If f(6) = 1.4 and f(8) = 1.7, then evaluate $\displaystyle \log_{3} 35$ and $\displaystyle \log_{3} (7/5)$

b. Find t such that f(t +2) + f(t + 4) = 1

2. Give an example of a logarithmic function g, which has a vertical asymptote x = –2, and passes through the origin.

2. Use your logarithm properties and function properties and make sense of it.

$\displaystyle f(6) = log_{3}(6-1) = log_{3}(5) = 1.4$

$\displaystyle f(8) = log_{3}(8-1) = log_{3}(7) = 1.7$

And finally

$\displaystyle log_{3}(35) = log_{3}(5\cdot 7) = log_{3}(5) + log_{3}(7) = 1.4 + 1.7$

You finish up and then do the other one.

3. Awesome! I got part a.

would part b look like this:

(t+2)(t+4) = 3

t^2+ 6t + 8 = 3
t^2 + 6t + 5 = 0
(t + 5) (t + 1)

= -5, -1, but I reject -5. so it's -1?

I don't know how to do part 2.

4. Originally Posted by goliath
(t+2)(t+4) = 3
You seem to know a few correct things, but you are forgetting the most fundamental. BE CAREFUL.

$\displaystyle f(t+2) = log_{3}((t+2)-1) = log_{3}(t+1)$

Now, would you like to rethink that Part B?

5. Ahh, that was my first guess, but i wasn't too sure.

$\displaystyle \log_{3} (t + 1) + \log_{3} (t + 3)$

(t+1)(t+3) = 3

then would i just do what i did before?

6. Yes, except that you can't factor it.

1) Don't write things that don't make sense. You should start with an equation. Your firt expression is not an equation.

2) How did you make the tansition from the equation you didn't write to the quadratic equation? There is a step missing.

3) You seem to be aware of Domain Considerations, but you did not state them when they were important. Before transforming to the quadratic equation, where there is no appearance of a Domain Restriction, WRITE DOWN the observable Domain Restrictions. In this case t > -3 and tell us why.

4) Generally, work so that you can follow it and so that every step and thought process can be reviewed.

7. Originally Posted by goliath
2. Give an example of a logarithmic function g, which has a vertical asymptote x = –2, and passes through the origin.
1) Here's a logarithmic function: $\displaystyle f(t) = b*log(x-a)$
1) Here's a logarithmic function: $\displaystyle g(t) = log_{b}(x-a)$
1) Here's a logarithmic function: $\displaystyle h(t) = b+log(x-a)$

2) Vertical asymptote x = –2: $\displaystyle f(t) = b*log(x+2)$
2) Vertical asymptote x = –2: $\displaystyle g(t) = log_{b}(x+2)$
2) Vertical asymptote x = –2: $\displaystyle h(t) = b+log(x+2)$

3) Passes through (0,0): $\displaystyle f(0) = b*log(0+2) = b*log(2) = 0$ -- No Good.
3) Passes through (0,0): $\displaystyle g(0) = log_{b}(0+2) = log_{b}(2) = 0$ -- No Good.
3) Passes through (0,0): $\displaystyle h(0) = b+log(0+2) = b+log(2) = 0$ -- Ah! This one can be solved for 'b'.

Very often, you just have to dive in.