# Thread: Transformations of Logarithmic Functions

1. ## Transformations of Logarithmic Functions

Hey guys
I have a question about transformations and the laws of logarithms.

The question is:
"Given y=1/5log(9x-36)^15-13 (the log has a base of 3, not 10. I don't know how to write that in clearly, so I hope it makes sense)
Apply the laws of logarithms to change the form of the equation. Graph the function by first stating the basic function and then describe each transformation applied in order. Specifically describe what happens to the domain, range, asymptotes, x-intercept, and vertical stretch or compression."

I am fine with graphing the transformation, I just don't understand exactly how to "apply the laws of logarithms to change the form of the equation"
Can anybody help me out a bit?

Thanks a lot

2. is the question?
$y=\frac{1}{5}\cdot \log_3 (9x-36)^{15}-13$

$y=\frac{1}{5}\cdot 15\log_3 (9x-36)-13$

$y=3\cdot\log_3 9(x-4)-13$

$y=3\cdot\log_3 9+\log_3(x-4)-13$

$y=3\cdot2 +\log_3(x-4)-13$

$y=\log_3(x-4)+6 -13$

$y=\log_3(x-4) -7$

....

lol i haven't done logs in a while.. only about 30% confident in my log abilities... i hope this isn't completely wrong and i'm steering you wrong.

3. Originally Posted by cazury
Hey guys
I have a question about transformations and the laws of logarithms.

The question is:
"Given y=1/5log(9x-36)^15-13 (the log has a base of 3, not 10. I don't know how to write that in clearly, so I hope it makes sense)
Apply the laws of logarithms to change the form of the equation. Graph the function by first stating the basic function and then describe each transformation applied in order. Specifically describe what happens to the domain, range, asymptotes, x-intercept, and vertical stretch or compression."

I am fine with graphing the transformation, I just don't understand exactly how to "apply the laws of logarithms to change the form of the equation"
Can anybody help me out a bit?

Thanks a lot

Let me make sure that I am reading your equation correctly first, is it

$y = \frac{1}{5} log_3 (9x-36)^{15} -13$

or

$y = \frac{1}{5 log_3 (9x-36)^{15}} -13$

4. I'm pretty sure that the exponent is not 15-13.

Really the only thing you can do to this equation, if the way I am reading it is correct is bring the 15 in front. You can't split up the (9x - 36) into two separate logs.:

$y = \frac{1}{5} log_3 (9x - 36)^{15} - 13$

$= 3 log_3 (9x - 36) - 13$

5. Originally Posted by ixo
is the question?
$y=\frac{1}{5}\cdot \log_3 (9x-36)^{15}-13$

$y=\frac{1}{5}\cdot 15\log_3 (9x-36)-13$

$y=3\cdot\log_3 9(x-4)-13$
$y=3(\log_3 9+\log_3(x-4))-13$

$y=3\cdot2 + 3\log_3(x-4)-13$

$y=3\log_3(x-4)+6 -13$

$y=3\log_3(x-4) -7$

6. Hey Guys
You got the equation right, sorry for not being more clear...
I was able to find an example of this question that I had missed when going over my course material. So you guys did get it right =)

Thanks so much for the help!

7. Originally Posted by mollymcf2009
I'm pretty sure that the exponent is not 15-13.

Really the only thing you can do to this equation, if the way I am reading it is correct is bring the 15 in front. You can't split up the (9x - 36) into two separate logs.:

$y = \frac{1}{5} log_3 (9x - 36)^{15} - 13$

$= 3 log_3 (9x - 36) - 13$
from $\log_3(9x-36)$ i'm factoring out $log_3[9(x-4)]$ and using $\log(a\cdot b)= \log(a)+ \log(b)$ to separate. Then using: If $x=b^y ,$ then $y = \log_b (x)$ to translate into $\log_3 9 = 2$ because $3^2 = 9$

Originally Posted by geld
$y=3(\log_3 9+\log_3(x-4))-13$

$y=3\cdot2 + 3\log_3(x-4)-13$

$y=3\log_3(x-4)+6 -13$

$y=3\log_3(x-4) -7$
Do i see a function in there? y=mx+b

Oh i see it.. i just added instead of factoring. Turns out my fundamental algebra is the prob not my log abilities. lol