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logx logx logx logx logx logx logx logx logx logx logx logx logx logx logx logx logx logx logx logxlogx logx logx logx logx logx
Our next step will be to include logarithms and trigonometric functions in our very well known one-variable equations.
Can you solve this equation?
log(x+21)+logx=2
If you need help, you can vist these links:
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I think the solutions are 4 and -25.
ReplyDeletelog(x+21)+logx=2
log(x^2+21x)=2
10^2=x^2+21x
x^2+21x-100=0
x=(-21+(441+400)^(1/2))/2=4
or
x=(-21-(441+400)^(1/2))/2=-25
But the second solution is not included on the real numbers.
I correct myself, both solutions are included in the real numbers. I was an stupid error.
ReplyDeleteYou are right; the solution is x=4.
DeleteThe second solution of the quadratic equation is not possible for equations including logarithms. It can not be a negative number.
Ok. That was the first that I thought, but then I realized that the minus disappears when you simplify:
ReplyDeletelog(-25+21)+log(-25)=
log(-4)+log(-25)=
log(-4*(-25))=
log100=2
So, I don't understand why can it be a solution.
That´s true. Althogh log(-25) doesn´t exist, in this situation, log(x+21)+logx=log(x+21)x; and the solutions are in (-∞,-21)U(0,∞); so x=-25 is a solution of log[(x+21)x]=2
ReplyDeleteI got the same answre as Nicolas, x1=4 and x2=-25, but i thought too that x2 didn't count as a solution, now I understand why it is a solution!
ReplyDeletex=-25 is not a solution of the equation log(x+21)+logx=2.
ReplyDeletex=-25 is a solution of log[(x+21)x]=2.