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Mathematics 13 Online
OpenStudy (anonymous):

How do you solve Log x (x+2)=2?

OpenStudy (anonymous):

x+2=x^2

OpenStudy (anonymous):

x^2-x-2=0 (x+1)(x-2)

OpenStudy (anonymous):

is it: \[\large \log _{10}x(x+2)=2\] ???

OpenStudy (anonymous):

x^2+2x=10^2?

OpenStudy (anonymous):

yeah that's what I thought but he's not answering.. :(

OpenStudy (anonymous):

o well i solved it the other way so you can solve it this way haha

OpenStudy (anonymous):

Thank you!

OpenStudy (anonymous):

ok

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