SOLUTION: I would greatly appreciate any help you can give me in working this problem -
Question: Use algebraic procedures to find the exact solutions of the equation.
Problem: log(2x^2+3x
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Exponential-and-logarithmic-functions
-> SOLUTION: I would greatly appreciate any help you can give me in working this problem -
Question: Use algebraic procedures to find the exact solutions of the equation.
Problem: log(2x^2+3x
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Question 108041: I would greatly appreciate any help you can give me in working this problem -
Question: Use algebraic procedures to find the exact solutions of the equation.
Problem: log(2x^2+3x)=log(10x+30)
Thank you for your assistance.
You can put this solution on YOUR website! Problem: log(2x^2+3x)=log(10x+30)
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Since the logs are equal, the anti-logs are equal:
(2x^2+3x)=(10x+30)
2x^2-7x-30=0
(x-6)(2x+5)=0
x = 6 or x = -5/2
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Need to check these "answers" in the original equation:
Checking x=6:
log(2*36+3*6) = log(60+30)
log(90) = log(90)
OK
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Checking x=-5/2
log(2*(25/4)+3*(-5/2)) = log(10*(-5/2)+3)
log((25/2)-(15/2)) = log(-25+3)
Won't work; logs of negatives don't exist.
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The only solution is x=6
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Cheers,
Stan H.
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