3log2x + 1/2log2x - 2log2(x+1)

I don't quite understand this. Ive gotten to "log2 x^3(x^1/2)/(x+1)^2"

Any help would be great.

Thanks.

~freak

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- Nov 30th 2009, 01:02 PMfreakWrite as a single logarithm.
3log2x + 1/2log2x - 2log2(x+1)

I don't quite understand this. Ive gotten to "log2 x^3(x^1/2)/(x+1)^2"

Any help would be great.

Thanks.

~freak - Nov 30th 2009, 01:43 PMqmechJoining logs
$\displaystyle

3log2x + (1/2)log2x - 2log2(x+1)

$

=$\displaystyle

log(2x)^3 + log(2x)^{1/2} - log{(2(x+1))}^2

$

=$\displaystyle

log(2x)^{3.5} - log{(2(x+1))}^2

$

=$\displaystyle

log \frac{(2x)^{3.5}}{{(2(x+1))}^2}

$ - Nov 30th 2009, 02:04 PMfreak
Thanks, i looked it over again and realized where i went wrong. I got the same asnwer as you did but the book had the top x to the power of 7/2. It took me to right after i posted this to realized that 3.5 and 7/2 were the same thing lol.

Thanks for the help.

~freak