SOLUTION: I just want to make sure that the answer I got for this problem is correct.
The problem is log 2-5x/2(x+8)= 0 I used the quotient rule and got log(2-5x)-log 2(x+8)=0. Then I adde
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-> SOLUTION: I just want to make sure that the answer I got for this problem is correct.
The problem is log 2-5x/2(x+8)= 0 I used the quotient rule and got log(2-5x)-log 2(x+8)=0. Then I adde
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Question 457422: I just want to make sure that the answer I got for this problem is correct.
The problem is log 2-5x/2(x+8)= 0 I used the quotient rule and got log(2-5x)-log 2(x+8)=0. Then I added log2(x+8)on the left to cancel it out and added it to the right side, but I also distributed it (I also dropped the logs). So I have (2-5x)= 2x+16. Then I solved for X and got 2. But I'm not sure if this is right. I'm confused as to how to enter it into my calculator so that the answer comes out right. Answer by bucky(2189) (Show Source):
You can put this solution on YOUR website! The way you have written your original problem is log 2 - 5x/2(x+8) = 0. Written in this way, the problem according to the rules of algebra should be interpreted as follows:
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However, this is not the problem that you worked. From what you worked you probably meant to write the problem in algebraic form as log((2-5x)/(2*(x+8))) and that is correctly interpreted as:
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Assuming that this second interpretation is correct, you proceeded correctly. Your methodology is correct down to the point where you had:
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Somewhere in here you made a minor sign error. Subtract 2 from both sides and you have:
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Then subtract 2x from both sides to get:
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Finally, divide both sides by -7 and you arrive at the answer:
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You indicated that your answer was 2 (presumed to be x = +2). That is perhaps why you are having trouble checking your answer with a calculator.
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You can return to your problem and substitute -2 for x and you will get:
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This is correct because and
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Hope this helps. Make sure that you fully understand how to write algebraic expressions in line form and that operations are performed from left to right within parentheses first, then exponents, then multiplication and divisions as encountered left to right and finally additions and subtractions left to right as encountered. It can make quite a difference in correctly interpreting the problem as you can see from the two forms above.