Question 1186990
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log2(x) + log2(x+1) = 1
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<pre>
I assume that the logarithms are base 2.


From

    log2(x) + log2(x+1) = 1


we have, using the property of logarithms,

    x*(x+1) = 2


Making FOIL, you get

    x^2 + x - 2 = 0.


Factor 

   (x+2)*(x-1) = 0.


We deny the negative root x= -2, since it does not work under logarithm.


The only remaining solution is x= 1.    <U>ANSWER</U>
</pre>

Solved.