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justme

  • 5 years ago

log (x-9) + log x = 1

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  1. radar
    • 5 years ago
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    \[10^{x-9}+10^{x}=10^{0}\] x-9+x=0 2x=9 x=4.5 check:\[10^{4.5-9}+10^{4.5}=10^{0}=1\]

  2. justme
    • 5 years ago
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    wow thanks! the way you wrote it made it so much easier to understand ( :

  3. amistre64
    • 5 years ago
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    log(x^2 - 9x) = 1 x^2 - 9x = 10^1 = 10 x^2 - 9x - 10 = 0 (x-10)(x+1) = 0 x = 10, -1 ...... log(10-9) + log(10) = 1 ?? log(1) + 1 = 1 log(1) = 0 10^0 = 1 ... did i do that right?

  4. amistre64
    • 5 years ago
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    log(-1-9) = log(-10)...-1 aint an option tho :)

  5. amistre64
    • 5 years ago
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    log(4.5 - 9) + log(4.5) = 1 log(-4.5) + log(4.5) logs cant be negative....

  6. amistre64
    • 5 years ago
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    bases dont add if they aint got the same exponents x^2 + x^3 doesnt equal x^5

  7. radar
    • 5 years ago
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    Can not you let them be negative to solve this problem?? lol

  8. amistre64
    • 5 years ago
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    well; we will have to write the board of logaritms and enter a plea, then the senate will ahve to vote on it .... its just alot of work for one litle problem :)

  9. justme
    • 5 years ago
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    base is 10 because it is not given. so all the bases here are the same.

  10. amistre64
    • 5 years ago
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    doesnt matter; bases dont add without like exponents x^2 + x^5 DOESNT EQUAL x^7

  11. amistre64
    • 5 years ago
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    10^a + 10^b DOESNT EQUAL 10^(a+b)

  12. amistre64
    • 5 years ago
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    logs arent numbers; so they dont work like numbers; they are exponents and follow the rules of exponents

  13. radar
    • 5 years ago
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    Yes, I do see the flaw in such an elegant solution, sorry about that justme, please ignore.

  14. justme
    • 5 years ago
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    so x=4.5 right? seems to work when checked.

  15. amistre64
    • 5 years ago
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    log(x-9) + log(x) = 1; exponents add when bases multiply log((x)(x-9)) = 1 log(x^2 -9x) = 1 10^log... = 10^1 x^2 -9x = 10 x^2 -9x -10 = 0 x= 10, -1 ; -1 is illegal to use in the equation becasue log(-1-9) doesnt exist

  16. radar
    • 5 years ago
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    adding as I did was really multiplying, and of course 10^x*10^-x would equal 1.

  17. amistre64
    • 5 years ago
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    4.5 is illegal to use; becasue log(-4.9) does not exist

  18. amistre64
    • 5 years ago
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    you can tell by graphing the log function and going over to -10 and -4.5 and looking around to find a y value.... it aint there :)

  19. radar
    • 5 years ago
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    You have convinced me this problem is above my pay grade, please post the correct solution if you can or anyone can solve it.

  20. amistre64
    • 5 years ago
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    the correct solution is: x = 10. And we all get lost in the math eventually ;)

  21. justme
    • 5 years ago
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    how did you get that?

  22. amistre64
    • 5 years ago
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    ........ if you scroll up and read the post, i actually typed it out a few times lol

  23. radar
    • 5 years ago
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    wouldn't that result in 11=1 we know thats not right.

  24. amistre64
    • 5 years ago
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    log(10-9) + log(10) = 1 log(1) + 1 = 1 log(1) = 0 10^log(1) = 10^0 1 = 10^0 1 = 1

  25. amistre64
    • 5 years ago
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    also known as: 0 + 1 = 1

  26. radar
    • 5 years ago
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    Thanks

  27. radar
    • 5 years ago
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    It will work.

  28. radar
    • 5 years ago
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    Justme, did you see the flaw in the way I worked it? I hope I didn't mess you up. But the addition was illegal, as we were dealing with exponents, adding is exponents is the same as multiplying, and your problem was addition. I learned a lot on that thing!!!

  29. justme
    • 5 years ago
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    radar, me too that was the best lesson on log I have seen. thanks!

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