Question 416707: Please help me solve this equation! log2 x+ log2 (x+1)=1
Found 3 solutions by anshu_a, MathTherapy, ikleyn:
Answer by anshu_a(16)   (Show Source):
Answer by MathTherapy(10806)  (Show Source):
You can put this solution on YOUR website!
Please help me solve this equation! log2 x+ log2 (x+1)=1
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 , with x > 0
The other person's claim, that - 2 is a solution to this equation, is FALSE.
1 is, though, since it's > 0!!
Answer by ikleyn(53750)  (Show Source):
You can put this solution on YOUR website! .
Please help me solve this equation! log2 x+ log2 (x+1)=1
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Below is my complete correct solution
log2(x) + log2(x+1) = 1
The domain for this equation is the set of all real positive numbers x > 0.
In this domain, the given equation is equivalent to this one
log2(x*(x+1)) = 1
Simplify and find x
x(x+1) = 2^1,
x^2+x-2 = 0
(x+2)(x-1) = 0
x = -2, 1
x = -2 is an extraneous solution out the domain, so we disregard x = -2
(since logarithm of the negative argument is not defined).
Thus, x = 1 is the unique real solution to the original equation. <<<---=== ANSWER
Solved.
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