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anonymous

  • one year ago

What is the solution of log4x − 1032 = 5?

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  1. anonymous
    • one year ago
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    \[\log_{4x-10} 32=5\]

  2. Nnesha
    • one year ago
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    change log to exponential form |dw:1433711587287:dw|

  3. anonymous
    • one year ago
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    \[4x-10^{32}=5\] like that?

  4. Nnesha
    • one year ago
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    nope let 5 = y so number which is at right side becomes exponent of base

  5. anonymous
    • one year ago
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    \[4x-10^{5}=\]

  6. anonymous
    • one year ago
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    =y i meant

  7. Nnesha
    • one year ago
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    not equal y y =5 x=32 b =4x-10

  8. Nnesha
    • one year ago
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    |dw:1433711954963:dw| it should equal to 32

  9. Nnesha
    • one year ago
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    \[(4x-10)^{5}=\color{red}{32}\] and don't forget the parnetheses

  10. Nnesha
    • one year ago
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    parentheses

  11. anonymous
    • one year ago
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    \[(4x-10)^{5}=32\]

  12. anonymous
    • one year ago
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    okay

  13. anonymous
    • one year ago
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    so now you solve or what?

  14. Nnesha
    • one year ago
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    yes solve that

  15. anonymous
    • one year ago
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    math is my weak spot soo yea

  16. Nnesha
    • one year ago
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    take 5th root both sides to cancel out 5

  17. Nnesha
    • one year ago
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    \[\huge\rm \sqrt[5]{(4x-10)^5} = \sqrt[5]{32}\] remember you can convert 5th root to 1/5 exponent and that's how you will cancel 5 exponent

  18. anonymous
    • one year ago
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    okay so that mean when you cancel out the 5 it will just be 4x-10?

  19. Nnesha
    • one year ago
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    yes that's right! same like u take square root to cancel out square now at right side write 32 in terms of 5 base ?^5 = 32 ?

  20. anonymous
    • one year ago
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    2?

  21. Nnesha
    • one year ago
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    yep right so \[\large\sqrt[5]{2^5} = 2^\frac{ 5 }{ 5 }=2\] now solve for x 4x-10 =2 simple algebra :-)

  22. anonymous
    • one year ago
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    x=3

  23. Nnesha
    • one year ago
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    yes that's right :-)

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