SOLUTION: log(x+3)+ logx =1 solve x and show that the above equation will have an irrational root if the base is changed to 2, and the a rational root if the base is changed to 4

Question 1116776: log(x+3)+ logx =1 solve x and show that the above equation will have an irrational root if the base is changed to 2, and the a rational root if the base is changed to 4
Answer by rothauserc(4718) (Show Source): You can put this solution on YOUR website!
case 1 is log(x+3)+ log (x) = 1 with base 2
:
(log(x)/log(2)) + (log(x+3)/log(2)) = 1
:
log(x) + log(x+3) = log(2)
:
log(x(x+3)) = log(2)
:
x(x+3) = 2
:
x^2 +3x = 2
:
complete the square
:
x^2 +3x +9/4 = 2 + 9/4 = 17/4
:
(x+(3/2))^2 = 17/4
:
x+(3/2) = square root(17)/2
:
x = square root(17)/2 -3/2
:
Note that this is the only solution for this case, the solution x = -square root(17)/2 -3/2 does not solve the equation(hint substitute for x)
:
case 2 is log(x+3)+ log (x) = 1 with base 4
:
x^2 +3x = 4
:
x^2 +3x -4 = 0
:
(x+4) * (x-1) = 0
:
x = -4, x = 1
:
x = 1 is the solution since logarithm of a negative number is a complex number
:

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