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Mathematics 7 Online
OpenStudy (anonymous):

Solve 4 log12 2 + log12 x = log12 96. x = 88 x = 80 x = 12 x = 6

OpenStudy (anonymous):

@IMStuck

OpenStudy (imstuck):

Ok the first thing you will do is move the log12 96 over to the other side by subtracting it. It looks like this to do that:

OpenStudy (imstuck):

\[4 \log _{12}2 + \log _{12}x - \log _{12}96=0\]

OpenStudy (imstuck):

First the exponent rule, even though there's only one exponent, the 4 in the first term. That moves like this, if you remember:

OpenStudy (imstuck):

\[\log _{12}(2)^{4}+\log _{12}x-\log _{12}96\]

OpenStudy (jdoe0001):

http://www.chilimath.com/algebra/advanced/log/images/466x428xrules,P20of,P20exponents.gif.pagespeed.ic.1z7-Ekl5L5.png <---- use the 3rd rule listed there then the 1st rule then move everything to the left side then use the 2nd rule

OpenStudy (imstuck):

ok it looks like someone else wants to take over, so ok...

OpenStudy (anonymous):

you can continue @IMStuck

OpenStudy (anonymous):

yes, we were

OpenStudy (imstuck):

Wow...this is an equation! Done totally different! Let me refresh...delete all that stuff up there, and start again.

OpenStudy (anonymous):

you'd have to delete it. i can only delete what i wrote

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