SOLUTION: Can anybody solve log2 x + log2(x − 2) = 3 those are log base 2
I tried following instructions and end up with {{{x^2-2x-9}}} because they say I'm supposed to square the
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-> SOLUTION: Can anybody solve log2 x + log2(x − 2) = 3 those are log base 2
I tried following instructions and end up with {{{x^2-2x-9}}} because they say I'm supposed to square the
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Question 806279: Can anybody solve log2 x + log2(x − 2) = 3 those are log base 2
I tried following instructions and end up with because they say I'm supposed to square the 3, but I don't know if that's right Found 3 solutions by solver91311, htmentor, jim_thompson5910:Answer by solver91311(24713) (Show Source):
Whoever "they" are, stop listening to them. (Or learn to listen better, because if "they" actually knew what they were talking about they most certainly did NOT tell you to square the 3.)
The sum of the logs is the log of the product, so
But
Hence
Factor and solve. Exclude any negative root (the domain of log is the positive reals).
John
Egw to Beta kai to Sigma
My calculator said it, I believe it, that settles it
You can put this solution on YOUR website! log(x) + log(x-2) = 3
Using the product rule we can write the LHS as log of the two factors:
log(x(x-2)) = 3
To remove the logarithm, raise the terms on either side of the equation to the 2nd power:
2^log(x(x-2) = 2^3
x^2 - 2x = 8
x^2 - 2x - 8 = 0
Factor:
(x-4)(x+2) = 0
This gives two solutions, x=4 and x=-2
But since we can't take the logarithm of a negative number, the answer is x=4.
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