You can put this solution on YOUR website! Solve for x: log20(x)+log20(x+1)=1
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adding logs means multiply so we can write it:
log20(x(x+1)) = 1
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Which is
log20(x^2+x) = 1
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Write the exponent equiv of logs
20^1 = (x^2 + x)
20 = x^2 + x
0 = x^2 + x - 20; a quadratic equation
Factor to
(x+5)(x-4) = 0
x= -5
x= +4, this is the only solution, can't have a log of a negative number
:
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Check solution x=4 in the original problem
log20(4)+log20(4+1)=1
log20(4)+log20(5)=1
log20(4*5)
log20(20) = 1, obviously
