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anonymous

  • 5 years ago

can any one solve this?(log(a)(x^3))+(log(a)(sqrtx))= answer has to be in klog(a)x form thanks already :)

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  1. amistre64
    • 5 years ago
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    is that base a?

  2. amistre64
    • 5 years ago
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    log[a](x^3) like this?

  3. amistre64
    • 5 years ago
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    3loga(x) + (1/2)loga(x)

  4. amistre64
    • 5 years ago
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    loga(x)(3+(1/2)) ??

  5. anonymous
    • 5 years ago
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    Amistre, I think contact has been lost lol.

  6. amistre64
    • 5 years ago
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    looks right lol (6/2) loga (x)

  7. amistre64
    • 5 years ago
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    :) but the math remains lol

  8. anonymous
    • 5 years ago
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    Hahahaa, so very true!

  9. amistre64
    • 5 years ago
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    (7/2) loga(x)

  10. anonymous
    • 5 years ago
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    yes a is the base, soz my internet played up :S

  11. amistre64
    • 5 years ago
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    can we factor out the loga(x) like that tho?

  12. anonymous
    • 5 years ago
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    \[\log_{a} x^3+\log_{a} sqrtx\]

  13. amistre64
    • 5 years ago
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    if we multiply it togeter again we get: loga[sqrt(x^7)] so its good

  14. amistre64
    • 5 years ago
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    (7/2) loga(x) is your answer

  15. anonymous
    • 5 years ago
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    where did 7 come from? soz but i dont get the stuff you lot did at the top :S

  16. amistre64
    • 5 years ago
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    3 + 1/2 = 3' 1/2 to get it into an improper fraction (topheavy) we multiply 3 by the denom and add the numerator; 3(2) = 6 + 1 = 7.... 7/2

  17. amistre64
    • 5 years ago
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    OR we could go the route of changing addition of logs back into multiplication of logs

  18. anonymous
    • 5 years ago
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    ok :) multiplication logs?

  19. amistre64
    • 5 years ago
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    loga(x^3 sqrt(x)) = loga(sqrt(x^7)) = loga(x^(7/2)) (7/2) loga(x)

  20. amistre64
    • 5 years ago
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    loga(x) + loga(y) = loga(xy)

  21. amistre64
    • 5 years ago
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    loga(x^n) = n loga(x) as well

  22. anonymous
    • 5 years ago
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    ok so 7/2logax is the answer right?

  23. amistre64
    • 5 years ago
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    yes; (7/2) loga(x) is the answer

  24. anonymous
    • 5 years ago
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    \[7/2\log_{a} x\]

  25. anonymous
    • 5 years ago
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    yay! ill give u all medals :) thx

  26. amistre64
    • 5 years ago
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    im gonna need a new cigar box to keep these things in ;)

  27. anonymous
    • 5 years ago
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    :p

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