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For 0 = -loge (x-2), solve for x. Would I bring over the negative sign first?


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1 Answer

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Best answer

Up to you. I would recommend it, since the 0 will absorb the negatives (-0 = 0)

0 = loge(x-2)
e^0 = x-2
1 = x-2
x = 3

Here's what happens if you don't do it:

0 = loge( (x-2)^(-1) )
0 = loge( 1/(x-2) )
e^0 = 1/(x-2)
1 = 1/(x-2)
(x-2) = 1
x = 3

A bit more complicated, but either way should make sense to you. Basic log laws in action.


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