# Thread: logarithm with a < 1

1. ## logarithm with a < 1

Hi;
I have the function f(x) = -1/2log(3x - 3) + 5.

I have a table of parent function values x:1/9,1/3,1,3,9 and y:-2,-1,0,1,2.

I'm ok with all transformations eccept when I have an a value less than 1.
I know its vertical compression by a factor of a but how do I apply it to my y values to get new values?

Thanks.

2. ## Re: logarithm with a < 1

Originally Posted by anthonye
Hi;
I have the function f(x) = -1/2log(3x - 3) + 5.

I have a table of parent function values x y
1/9 -2
1/3 -2
1 0
3 1
9 2

I'm ok with all transformations eccept when I have an a value less than 1.
I know its vertical compression by a factor of a but how do I apply it to my y values to get new values?

Thanks.
I need some clarification. In your table are you stating that x = 1/9 gives you a y value of -2? If so log(3 (1/9) - 3) = log(1/3 - 3) = log(-8/3) which is not a real number.

-Dan

3. ## Re: logarithm with a < 1

Oh I think I see, If I have an x value<1 and a<1 do I always end up taking the log of a negative?

4. ## Re: logarithm with a < 1

Originally Posted by anthonye
Oh I think I see, If I have an x value<1 and a<1 do I always end up taking the log of a negative?
What's a?

The argument of the log has to be a positive number. So your condition is $\displaystyle 3x - 3 > 0$. So the only permitted x value in your list are 3 and 9. There is a problem with your y values though. For example:
$\displaystyle y(3) = -\frac{1}{2} log(3 \cdot 3 - 3) + 5= -\frac{1}{2} log(6) + 5 \neq 1$

-Dan

5. ## Re: logarithm with a < 1

Originally Posted by anthonye
Hi;
I have the function f(x) = -1/2log(3x - 3) + 5.

I have a table of parent function values x:1/9,1/3,1,3,9 and y:-2,-1,0,1,2.

I'm ok with all transformations eccept when I have an a value less than 1.
I know its vertical compression by a factor of a but how do I apply it to my y values to get new values?

Thanks.
given the table values, it looks as though your parent function is $\displaystyle y = \log_3{x}$ , a base 3 log.

now, if f(x) is also a base 3 log function (I'm just guessing here ... )

$\displaystyle f(x) = -\frac{1}{2} \log_3(3x-3) + 5$

$\displaystyle f(x) = -\frac{1}{2} \log_3[3(x-1)] + )5$

$\displaystyle f(x) = -\frac{1}{2} \left[\log_3{3} + \log_3(x-1) \right] + 5$

$\displaystyle f(x) = -\frac{1}{2} \left[1 + \log_3(x-1) \right] + 5$

$\displaystyle f(x) = -\frac{1}{2} - \frac{1}{2} \log_3(x-1) + 5$

$\displaystyle f(x) = -\frac{1}{2} \log_3(x-1) + \frac{9}{2}$

best I can make out, it looks like f(x) is the parent function transformed as follows ...

a horizontal shift right one unit, a vertical compression by a factor of 1/2, a reflection over the x-axis, and finally a vertical shift up 4.5 units

note the transformed function has domain x > 1

6. ## Re: logarithm with a < 1

I have a table of parent function values x:1/9,1/3,1,3,9 and y:-2,-1,0,1,2.
transformed x-values (all shifted right one unit) ...

10/9 , 4/3 , 2 , 4 , 10

transformed y-values (after the stated compression, reflection, and vertical shift) ...

5.5 , 5 , 4.5 , 4 , 3.5

7. ## Re: logarithm with a < 1

Yes parent function f(x) = log3(x) '' that 3 is the base''

skeeter that still gives me a negative in the log so how did you get the values?

Thanks.

8. ## Re: logarithm with a < 1

All x-values shift to the right one unit ... look at post #6

9. ## Re: logarithm with a < 1

Yes just realized and log on.

Thanks again.