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log(x+21)+logx=2

 Jan 10, 2016
 #1
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Solve for x:
log(x)+log(x+21) = 2

log(x)+log(x+21) = log(x (x+21)):
log(x (x+21)) = 2

Cancel logarithms by taking exp of both sides:
x (x+21) = e^2

Expand out terms of the left hand side:
x^2+21 x = e^2

Add 441/4 to both sides:
x^2+21 x+441/4 = e^2+441/4

Write the left hand side as a square:
(x+21/2)^2 = e^2+441/4

Take the square root of both sides:
x+21/2 = sqrt(e^2+441/4) or x+21/2 = -sqrt(e^2+441/4)

Subtract 21/2 from both sides:
x = sqrt(e^2+441/4)-21/2 or x+21/2 = -sqrt(e^2+441/4)

Subtract 21/2 from both sides:
x = sqrt(e^2+441/4)-21/2 or x = -21/2-sqrt(e^2+441/4)

log(x)+log(x+21) => log(-21/2-sqrt(e^2+441/4))+log(21+(-21/2-sqrt(e^2+441/4)))  =  2+(2 i) pi ~~ 2.+6.28319 i:
So this solution is incorrect

log(x)+log(x+21) => log(21+(sqrt(e^2+441/4)-21/2))+log(sqrt(e^2+441/4)-21/2)  =  2:
So this solution is correct

The solution is:
Answer: | x = sqrt(e^2+441/4)-21/2=0.346153977282...........

 Jan 10, 2016

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