log(x-1) - log(x+6) = log(x-2) - log(x-3) Is this right? (x-1)/(x+6) = (x-2)/(x-3) (x-3)(x-1)/(x+3)(x+6) = (x-2)(x+6)/(x+3)(x+6) (x-1)/(x-6) = (x-2)/(x-3) x^2 + 4x - 12 = x^2 - 4x - 3 x = 9/8
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Looks right. BUT log(x-2)=log(9/8-2)=log(-7/8) which is not defined. So, something's wrong ! -O
Originally Posted by NeedHelp18 log(x-1) - log(x+6) = log(x-2) - log(x-3) Is this right? (x-1)/(x+6) = (x-2)/(x-3) (x-3)(x-1)/(x+3)(x+6) = (x-2)(x+6)/(x+3)(x+6) (x-1)/(x-6) = (x-2)/(x-3) huh? shouldn't that line be (x - 1)(x - 3) = (x - 2)(x + 6)? ...(which would yield (x - 1)/(x + 6) = (x - 2)/(x - 3) ...but that's not what we would do) expand both sides: now solve
Hello, NeedHelp18! Is this right? . . . . What was all that about? . . . . no We have: . Rearrange terms: . And we have: . Exponentiate: . . . which simplifies to: . But this answer turns out to be extraneous. The equation has no solution.
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