Solve 4 log12 2 + log12 x = log12 96. x = 88 x = 80 x = 12 x = 6
@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:
\[4 \log _{12}2 + \log _{12}x - \log _{12}96=0\]
First the exponent rule, even though there's only one exponent, the 4 in the first term. That moves like this, if you remember:
\[\log _{12}(2)^{4}+\log _{12}x-\log _{12}96\]
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
ok it looks like someone else wants to take over, so ok...
you can continue @IMStuck
yes, we were
Wow...this is an equation! Done totally different! Let me refresh...delete all that stuff up there, and start again.
you'd have to delete it. i can only delete what i wrote
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