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anonymous

  • 5 years ago

Log4 X-log16 (x+3)=1/2 The 4 and 16 are bases can someone please tell me how to solve this?

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  1. amistre64
    • 5 years ago
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    change of base perhaps??

  2. amistre64
    • 5 years ago
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    4^y = x 16^y-3 = x would that work out ?

  3. anonymous
    • 5 years ago
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    change both bases to 4, then use the laws of exponents to combine\[\log_{16}(x+3)=y\] means \[16^y=(x+3)\] \[4^{2y}=x+3\] \[2y=log_4(x+3)\] \[y=\frac{log_4(x+3)}{2}\]

  4. amistre64
    • 5 years ago
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    4^y - 4^(2y-6) = 1/2

  5. amistre64
    • 5 years ago
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    i was close lol

  6. anonymous
    • 5 years ago
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    change of base gets me two joined equations with a factored out x

  7. amistre64
    • 5 years ago
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    log4(x) - [log4(x+3)/log4(16)] = 1/2 log4(x) - (1/2)log4(x+3) = 1/2

  8. anonymous
    • 5 years ago
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    idk if i can change both bases to 4

  9. amistre64
    • 5 years ago
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    log4(x/sqrt(x+3)) = 1/2

  10. anonymous
    • 5 years ago
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    you can do that?

  11. amistre64
    • 5 years ago
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    sqrt(4) = sqrt(x/(x+3)) i believe so :)

  12. anonymous
    • 5 years ago
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    then exponentiate to get \[\frac{x}{\sqrt{x+3}}=4^{\frac{1}{2}}\]

  13. amistre64
    • 5 years ago
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    4 = x/x+3 4x+12 = x 3x = -12 x = -4...maybe

  14. amistre64
    • 5 years ago
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    i got lost in it for a moment....

  15. anonymous
    • 5 years ago
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    no way -4 is an answer since you cannot take log of -4

  16. amistre64
    • 5 years ago
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    i know i know.... my brain took a momentary leave of absence lol

  17. anonymous
    • 5 years ago
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    the 4 is not negative

  18. amistre64
    • 5 years ago
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    sqrt(4) = x/(x+3) is what i meant to type :)

  19. amistre64
    • 5 years ago
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    4 = x^2/x^2+9+6x

  20. amistre64
    • 5 years ago
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    4x^2 +24x +36 = x^2 3x^2 +24x +36 = 0... thats better me thinks

  21. amistre64
    • 5 years ago
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    3(x^2 +8x +12) = 0

  22. amistre64
    • 5 years ago
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    x = -6 and -2 by that accounting...... so maybe not ;)

  23. anonymous
    • 5 years ago
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    oops start with \[\frac{x}{\sqrt{x+3}}=4^{\frac{1}{2}}\]

  24. amistre64
    • 5 years ago
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    ...... yeah, it would help if I did the actual problem instead of making up me own lol

  25. anonymous
    • 5 years ago
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    \[x=2\sqrt{x+3}\]

  26. anonymous
    • 5 years ago
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    \[x^2=4(x+3)=4x+12\] \[x^2-4x-12=0\] \[(x-6)(x+2)=0\] \[x=6, x=-2\] -2 not a solution. we can try 6 to see if it works.

  27. anonymous
    • 5 years ago
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    yup 6 it is

  28. anonymous
    • 5 years ago
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    hey where did the log16 go?

  29. amistre64
    • 5 years ago
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    we..and be that i me sat.... modified it to a log4(x+3)/2

  30. anonymous
    • 5 years ago
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    I get it its just i do not know where the log16 went the log of16 is 1.20411

  31. amistre64
    • 5 years ago
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    change of base just changes the appearance and not the value

  32. amistre64
    • 5 years ago
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    log4(16) not log(16)

  33. anonymous
    • 5 years ago
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    let us go slow. \[log_{16}(x)=y\] means \[16^y=x\] for example \[log_{16}(4096)=3\] because \[16^3=4096\]

  34. anonymous
    • 5 years ago
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    but \[16=4^2\] so it is also true that \[(4^2)^3=4069\] \[4^6=4069\] so \[log_4(4069)=6\]

  35. anonymous
    • 5 years ago
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    so the log base 16 did not disappear, just converted to log base 4

  36. anonymous
    • 5 years ago
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    oho i get it thank you :]

  37. anonymous
    • 5 years ago
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    i got X=2.6667

  38. anonymous
    • 5 years ago
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    how did you get that? i am fairly certain x = 6

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