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Mathematics 8 Online
OpenStudy (gabrielah96):

solve for x. log(x+12)=2log(x)

OpenStudy (campbell_st):

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

OpenStudy (lochana):

that's right now you can factorize it easily x^2 -x -12 = (x - 4)(x + 3) x = 4 and -3

ganeshie8 (ganeshie8):

reject -3 because we need `x > 0` for the expression `log(x)` to be real

OpenStudy (campbell_st):

are you sure about x = -3 log (-3 + 12) = log((-3)^2) I think you have to include x = -3 as a solution

OpenStudy (campbell_st):

it's always an interesting question...

ganeshie8 (ganeshie8):

\(\log(x^2)\) is not necessarily same as \(2\log x\)

ganeshie8 (ganeshie8):

that holds only when \(x\) is positive

ganeshie8 (ganeshie8):

OpenStudy (gabrielah96):

thanks everyone!

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