Question 277250
logx=1-log(x-3)
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applying the "log rules":
logx=1-log(x-3)
logx+log(x-3) = 1
log[x(x-3)] = 1
x(x-3) = 10^1
x^2-3x = 10
x^2-3x-10 = 0
factoring
(x-5)(x+2) = 0
x = {-2,5}
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We can toss out the -2 -- because plugging it back into the original eq we will get a log of a neg number -- no good -- this is called an "extraneous solution".
.
So, we conclude:
x = 5