Attached is the question and possible solution. Is it correct?
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$\displaystyle log(x+99)-log(3x-2)=2$ Rewrite: $\displaystyle log\left(\frac{x+99}{3x-2}\right)=2$ $\displaystyle \frac{x+99}{3x-2}=100$ $\displaystyle x+99=300x-200$ $\displaystyle -299x=-299$ $\displaystyle x=1$
It is not correct. Should be : log(x+99) - log(3x-2) = 2 log( x+99 / 3x-2 ) = 2 x+99 / 3x-2 = 10^2 x+99 = 100(3x-2) x+99 = 300x - 200 299x = 299 x = 1
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