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animalsavior94

  • 4 years ago

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

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  1. nowhereman
    • 4 years ago
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    So how far do you get?

  2. animalsavior94
    • 4 years ago
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    not at all because its so confusing

  3. anonymous
    • 4 years ago
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    is this log base 5?

  4. anonymous
    • 4 years ago
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    actually it doesnt matter so lets just write log

  5. anonymous
    • 4 years ago
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    and ignore the 1/3 out front we will deal with it last

  6. animalsavior94
    • 4 years ago
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    yes its a base 5

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

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

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

  10. anonymous
    • 4 years ago
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    now we use \[\log(a)-log(b)=log(\frac{a}{b})\]

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

  12. anonymous
    • 4 years ago
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    i will deal with that last

  13. anonymous
    • 4 years ago
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    not done yet!

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

  15. anonymous
    • 4 years ago
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    now we use \[\log(a)+\log(b)=\log(ab)\]

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

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

  18. anonymous
    • 4 years ago
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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 \[\log(\frac{x^2(x+1)^{\frac{4}{3}}}{3-x})\]

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

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