# Logarithmic Relationships

• May 20th 2012, 04:45 PM
strdatmage
Logarithmic Relationships
How would i solve this?

Log3^9-log3^27+log3^243
• May 20th 2012, 04:53 PM
Plato
Re: Logarithmic Relationships
Quote:

Originally Posted by strdatmage
How would i solve this?
Log3^9-log3^27+log3^243

There is nothing there to solve.
If you want to add those up then:
$9L+27L+243L=~?$ where $L=\log(3)$
• May 20th 2012, 04:56 PM
strdatmage
Re: Logarithmic Relationships
Quote:

Originally Posted by Plato
There is nothing there to solve.
If you want to add those up then:
$9L+27L+243L=~?$ where $L=\log(3)$

It says : Write each expression as a single logarithm. B. Find the value of each expression
• May 20th 2012, 05:37 PM
Plato
Re: Logarithmic Relationships
Quote:

Originally Posted by strdatmage
It says : Write each expression as a single logarithm. B. Find the value of each expression

That is exactly what I posted.
Now you do some work for yourself.
• May 20th 2012, 05:42 PM
strdatmage
Re: Logarithmic Relationships
Okay, so just Do that and replace L With 3?
• May 20th 2012, 06:03 PM
Reckoner
Re: Logarithmic Relationships
Quote:

Originally Posted by strdatmage
It says : Write each expression as a single logarithm. B. Find the value of each expression

As Plato demonstrated, you can use this property of logarithms:

$\log_bx^a=a\log_bx,\quad x>0, b>0, b\ne1$.

Then combine like terms. You should familiarize yourself with basic logarithmic identities. Consult your textbook for information.
• May 20th 2012, 11:14 PM
earboth
Re: Logarithmic Relationships
Quote:

Originally Posted by strdatmage
How would i solve this?

Log3^9-log3^27+log3^243

I'm a little bit confused about the writing of this sum ...

Could it be that the original term was:

$\log_3(9) - \log_3(27) + \log_3(243)$

If so the final result is 4.
• May 22nd 2012, 01:15 PM
ineedhelplz
Re: Logarithmic Relationships
Quote:

Originally Posted by strdatmage
It says : Write each expression as a single logarithm. B. Find the value of each expression

9L + 27L + 243L = 279L; where L = log(3)

279log(3) is the single logarithm
Use your calculator to find the value the appropriate base as it wasn't detailed here (assuming base e or base 10)