Question 371941: Solve the following logarithmic equation. Check all solutions.
log(x)+log(x+21)=2 Found 3 solutions by nerdybill, CharlesG2, Fombitz:Answer by nerdybill(7384) (Show Source):
You can put this solution on YOUR website! log(x)+log(x+21)=2
log(x(x+21))=2
x(x+21) = 10^2
x^2+21x = 100
x^2+21x-100 = 0
Factoring the left:
(x+25)(x-4) = 0
.
x = {-25, 4}
We can throw out the -25 solution -- it is an extraneous solution leaving:
x = 4
log(x) + log(x + 21) = 2
logarithmic rule: logb(m) + logb(n) = logb(mn)
log(x(x + 21)) = 2
log when no base is noted is assumed to be base 10
logarithmic rule: if logb(y) = x then b^x = y
10^2 = x(x + 21)
100 = x^2 + 21x
0 = x^2 + 21x - 100
0 = (x + 25)(x - 4)
by FOIL this is x^2 - 4x + 25x - 100 or x^2 + 21x - 100
x = -25 or x = 4
can not take log of a negative number
10 to the something can not equal -25
only solution is x = 4
check: log(4) + log(4 + 21) = 2 = log(4) + log(25)
log(4) + log(25) = log(100) = 2
10^2 = 100, yes
(