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log10x+log10(x+3)=1
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hint: \[Log(a)+Log(b)=Log(a*b) \]
so log (x^2 + 3x) = 10?
i mean = 1?
yes
so how would you solve for x ?
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x^2+3x=10^1
because \[10^{\log_{10}(x^2+3x) }=10^1\] and \[10^{\log_{10}(x^2+3x) }=x^2+3x\]
do you understand?
yes i think so
do the x's equal -5 and 2
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i put that as the answer but the computer's saying that it's incorrect
that's because you have to check if the answers are possible. In this case, when you fill in x=-5 You'll get Log10(-5)+Log10(-2)=1
you can't take the Log of a negative number. So the only right answer is x=2
oh ok that makes so much sense
thank you!
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