# Logarithm question....need help

• Dec 13th 2012, 03:26 PM
koolaid123
Logarithm question....need help
Write http://www.mathway.com/math_image.as...%29?p=124?p=42as a sing logarithm...x>0
i simplified it by moving the exponents to the side and i got to this http://www.mathway.com/math_image.as...%29?p=294?p=42
i dont know what to do from here......plz help
• Dec 13th 2012, 03:40 PM
Plato
Re: Logarithm question....need help
Quote:

Originally Posted by koolaid123
Write http://www.mathway.com/math_image.as...%29?p=124?p=42as a sing logarithm...x>0
i simplified it by moving the exponents to the side and i got to this http://www.mathway.com/math_image.as...%29?p=294?p=42
i dont know what to do from here......plz help

You have one mistake.
It is $\displaystyle -4$ not $\displaystyle +4\log_a$.

It is not clear what more you what to be done?
• Dec 13th 2012, 03:43 PM
koolaid123
Re: Logarithm question....need help
its still positive 4....i forgot to put brackets before the 3 and and the end....those two are multiplied by each other.....i have to write this as a single logarithm
• Dec 13th 2012, 05:42 PM
Plato
Re: Logarithm question....need help
Quote:

Originally Posted by koolaid123
its still positive 4....i forgot to put brackets before the 3 and and the end....those two are multiplied by each other.....i have to write this as a single logarithm

You are wrong.
$\displaystyle \log _a \left( {\frac{{\sqrt {x^2 + 1} }}{{x^3 \left( {x + 1} \right)^4 }}} \right) = \frac{1}{2}\log _a \left( {x^2 + 1} \right) - 3\log _a (x) - 4\log_a \left( {x + 1} \right)$