SOLUTION: solve: e^(x+1) = 20 log(x^2 + 19) = 2 logx + log(x+3) = 1

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: solve: e^(x+1) = 20 log(x^2 + 19) = 2 logx + log(x+3) = 1      Log On


   

Question 622918: solve:
e^(x+1) = 20
log(x^2 + 19) = 2
logx + log(x+3) = 1
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
solve:
e^(x+1) = 20
(x+1)lne=ln20
x+1=ln20
x=ln20-1
..
log(x^2 + 19) = 2
10^2=x^2+19
x^2=100-19
x^2=81
x=±9
..
logx + log(x+3) = 1
log[x(x+3)]=1
10^1=x^2+3x
x^2+3x-10=0
(x+5)(x-2)=0
x=-5(reject,x>0)
or
x=2