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Mathematics
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solve for x. log(x+12)=2log(x)
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well you can rewrite them using log laws \[lof(x +12) = \log(x^2)\] then raise each side as a power of the base and you then need to solve x + 12 = x^2 or x^2 - x - 12 = 0 now solve for x. hope it helps
that's right now you can factorize it easily x^2 -x -12 = (x - 4)(x + 3) x = 4 and -3
reject -3 because we need `x > 0` for the expression `log(x)` to be real
are you sure about x = -3 log (-3 + 12) = log((-3)^2) I think you have to include x = -3 as a solution
it's always an interesting question...
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\(\log(x^2)\) is not necessarily same as \(2\log x\)
that holds only when \(x\) is positive
thanks everyone!
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