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solve for x : log(3x+7)+log(x-2)=1show work plzz....i need to understand how to do this
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are these log base e or 10?
Log base 10 makes more sense. Combine logs using the property that log(a)+log(b)=log(ab) \[\log((3x+7)(x-2))=1\] Raise everything by 10. \[(3x+7)(x-2)=10^1\] So: \[3x^2+7x-6x-14=10\] \[3x^2+x-24=0\] Which factors to: \[(3x-8)(x+3)=0\] So x=8/3 and -3
thanks a bunch
No problem :P
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