# One more question

• Mar 3rd 2014, 02:53 PM
Rayzala
One more question
Just a couple more im confused about

I have one question that asks me to 'apply the laws of logarithms' to change the form of the following equation

Attachment 30284

Once again the samples are way easier and dont prepare me I need to change the form so that I can describe the transformations
the result I got was

3log_3(x-4)-1

It feels as if I'm doing it wrong though could you please show me how to get the final form so I can do the transformations?
• Mar 3rd 2014, 03:07 PM
Plato
Re: One more question
Quote:

Originally Posted by Rayzala
Attachment 30284
Once again the samples are way easier and dont prepare me I need to change the form so that I can describe the transformations
the result I got was 3log_3(x-4)-1

I get $3\log_3(x-4)-7$ Recall $3\log_3(9)=6$
• Mar 3rd 2014, 04:19 PM
Rayzala
Re: One more question
Quote:

Originally Posted by Plato
I get $3\log_3(x-4)-7$ Recall $3\log_3(9)=6$

So to find the transformations I would use the first function you gave me right?

Would the description of the transformation be the following?:

The base function is y=log_3x, stretched vertically by 4, and translated right by 4 and down by 13.
• Mar 3rd 2014, 04:59 PM
Plato
Re: One more question
Quote:

Originally Posted by Rayzala
So to find the transformations I would use the first function you gave me right?
Would the description of the transformation be the following?:
The base function is y=log_3x, stretched vertically by 4, and translated right by 4 and down by 13.

OH! come on! Learn the basics.

$\dfrac{1}{5}\log_3(9x-36)^{15}-13\\3\left[\log_3(9(x-4))\right]-13\\3\left[2+\log_3(x-4)\right]-13\\6+3\left[\log_3(x-4)\right]-13$
• Mar 3rd 2014, 05:10 PM
Rayzala
Re: One more question
Stretched vertically by 5, and translated right by 4 and down by 13?