Question 201660This question is from textbook
: log(square root of 1-x) - log(square root of x+2)=logx
First I sqared both sides and got log 1-x -log x+2=log x squared
then i performed the quotient property of logs and got log(1-x/x+2)=log x squared then I got 1-x =x squared (x-2) then I got X cubed - 2x+ x-1=0
This question is from textbook
Found 3 solutions by RAY100, stanbon, solver91311:
Answer by RAY100(1637)   (Show Source):
You can put this solution on YOUR website! Good work,,,,but watch signs for last 2 steps
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log{ (1-x) / (x+2) } = log (x^2)
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(1-x) / (x+2) = x^2
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(1-x) = (x^2) (x+2)
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1-x = x^3 + 2x^2
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0 = x^3 +2x^2 +x -1
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! log(square root of 1-x) - log(square root of x+2)=logx
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log[sqrt(1-x)/sqrt(x+2)] = log(x)
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sqrt[(1-x)/(x+2)] = x
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Square both sides:
(1-x)/(x+2) = x^2
(1-x) = x^3 + 2x^2
Rearrange:
x^3 + 2x^2 + x -1 = 0
x = 0.46557123...
====================
Cheers,
Stan H.
Answer by solver91311(24713)  (Show Source):
You can put this solution on YOUR website!
In the first place, you didn't actually square both sides. Squaring the log function does NOT result in the log of the square of the argument. Furthermore, you are trying to say that ^2 = a^2 + b^2) -- a statement that is false in general. Had you actually squared both sides of your original equation you would have a result that looked like:
 - \log(\sqrt{1 - x})\cdot\log(\sqrt{x + 2}) + \log^2(\sqrt{x + 2}) = \log^2(x))
I used to think that a mud fence was the gold standard for ugliness, but looking at that horror, I may change my mind.
Having said all of that, you actually stepped in a big pile of it and still came up smelling like a rose.
Remember your laws of logarithms. First:
 = n\log_b(x))
So back to the original equation:
 - \log(\sqrt{x + 2}) = log(x))
Which, recalling that  , can be re-written:
 - \frac{1}{2}\log(x + 2) = \log(x))
Multiplying both sides by 2:
 - \log(x + 2) = 2\log(x) = \log(x^2))
Which, coincidentally, and only coincidentally, is where you were after you say you squared both sides. Like I said, you were lucky.
From there you correctly applied the rule that the difference of the logs is the log of the quotient:
 = \log(x^2))
And then you correctly asserted that if two logs to the same base are equal, then their arguments must be equal:
From here you had a small error. Somehow in multiplying both sides by  , you ended up with  on the right. You should have arrived at:
)
And arranging into standard form results in:
Applying the rational root theorem, the only possible rational roots are 
I'll leave it for you to verify, but there are no rational roots.
That leaves us with applying the general solution of the cubic.
Fortunately, this cubic is monadic, so we can use a simplified form of the general solution:
For the cubic equation 
Where
The expression above for  can generate up to three values (there are three cubic roots related by a factor which is one of the two complex cubic roots of one, and two square roots of any sign ; but these 6 expressions can generate only 3 pairs). This also applies to the final solutions for  .
Lucky you -- you get to do your own arithmetic.
John
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