solve for x: logx+log(x+21)=2

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solve for x: logx+log(x+21)=2

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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the answer is 4 I just don't know how to get there
log(y)+log(z)=log(yz) tried to used that here
log[x(x+21)]=2

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Other answers:

alexa have you made an attempt to solve the question?
yes i have and I come up with something completely different. im not sure whether to use the quotient, product, or power rule
combine the product and quotient rule
to get rid of the log write base 10 on both sides so we have x(x+21)=10^2
log(x(x+21)=2 logx^2+21x=2 =>10^2=x^2+21
ummmm 21x
100=x^2+21x
right
simple quadratics
alexa I hope its alright now... problem solved?
I can go from here, thanks
you're welcome :)
heyy btw alexa, you got msn or fb or some personal contact we can talk later, I'd appreciate it :/
hey alex make sure you check your answers because we can't have log(negative)
sorry but I dont like to give my personal information out unless I personally know the individual, no offense. and thanks, I will.
np, understood :)

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