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animalsavior94 Group Title

write as logarithm of a single quantity: 1/3[4log5(x+1)-3log5(3-x)+6log5x]

  • 3 years ago
  • 3 years ago

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  1. nowhereman Group Title
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    So how far do you get?

    • 3 years ago
  2. animalsavior94 Group Title
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    not at all because its so confusing

    • 3 years ago
  3. satellite73 Group Title
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    is this log base 5?

    • 3 years ago
  4. satellite73 Group Title
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    actually it doesnt matter so lets just write log

    • 3 years ago
  5. satellite73 Group Title
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    and ignore the 1/3 out front we will deal with it last

    • 3 years ago
  6. animalsavior94 Group Title
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    yes its a base 5

    • 3 years ago
  7. satellite73 Group Title
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    first use \[\log(a^n)=n\log(a)\] backwards. all the multipliers come inside the log as exponents

    • 3 years ago
  8. satellite73 Group Title
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    \[4\log(x+1)=\log((x+1)^4)\] \[3\log(3-x)=\log((3-x)^3)\] \[6\log(x)=\log(x^6)\]

    • 3 years ago
  9. satellite73 Group Title
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    giving \[\log((x+1)^4)-\log((3-x)^3)+\log(x^6)\]

    • 3 years ago
  10. satellite73 Group Title
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    now we use \[\log(a)-log(b)=log(\frac{a}{b})\]

    • 3 years ago
  11. animalsavior94 Group Title
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    what happened to the 1/3 outside the []

    • 3 years ago
  12. satellite73 Group Title
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    i will deal with that last

    • 3 years ago
  13. satellite73 Group Title
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    not done yet!

    • 3 years ago
  14. satellite73 Group Title
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    \[log((x+1)^4)-\log((3-x)^3)=\log(\frac{(x+1)^4}{(3-x)^3})\]

    • 3 years ago
  15. satellite73 Group Title
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    now we use \[\log(a)+\log(b)=\log(ab)\]

    • 3 years ago
  16. satellite73 Group Title
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    \[\log(\frac{(x+1)^4}{(3-x)^3})+\log(x^6)=\log(\frac{x^6(x+1)^4}{(3-x)^3})\]

    • 3 years ago
  17. satellite73 Group Title
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    now for the 1/3 out front. take the cube root of all that stuff.

    • 3 years ago
  18. satellite73 Group Title
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    the inner stuff i mean. the cube root of x^6 is x^2, and the cube root of (3-x)^3 is 3-x and so you get \[\log(\frac{x^2(x+1)^{\frac{4}{3}}}{3-x})\]

    • 3 years ago
  19. animalsavior94 Group Title
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    wow that was a hard problem but thanks again:)

    • 3 years ago
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