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JJenk

  • one year ago

Solve. open parentheses square root of 7 close parentheses to the 6 x power = 49x-6

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  1. marissalovescats
    • one year ago
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    Can you draw this please?

  2. marissalovescats
    • one year ago
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    |dw:1373139028363:dw|

  3. JJenk
    • one year ago
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    |dw:1373139009699:dw|

  4. marissalovescats
    • one year ago
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    Hm... I recognize this. Let me go look at past notes.

  5. JJenk
    • one year ago
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    These were the choices i was given x = negative 21 over 2 x = -6 x = negative 6 over 5 x = -12

  6. marissalovescats
    • one year ago
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    Does this happen to be from a lesson called "Solve Exponential Equations" ?

  7. JJenk
    • one year ago
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    Yes

  8. marissalovescats
    • one year ago
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    Okay.. Uhm.. I've never seen one with a square root but let me see if I can solve it then I'll show you

  9. Jhannybean
    • one year ago
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    \[\large \sqrt{(7)}^{6x}=49^{x-6}\]\[\large (\sqrt{7})^{6x} = \color{green}{(\sqrt{7})^{4(x-6)}}\]\[\large 6x = 4(x-6)\]\[\large 6x=4x-24\] solve for x.

  10. marissalovescats
    • one year ago
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    No^^^ Its not 7^4. It's 7^2.

  11. marissalovescats
    • one year ago
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    I got -6/5 and I'll show you.

  12. marissalovescats
    • one year ago
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    |dw:1373139792425:dw|

  13. Jhannybean
    • one year ago
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    i changed it to the sae base.\(\sqrt{7}^4\)=49. and you've gotthe base \(\sqrt{7}\) on the left side.

  14. Jhannybean
    • one year ago
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    Tell me how you changed \(\large (\sqrt{7})^{6x}\) to just 7?

  15. marissalovescats
    • one year ago
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    I just figured because you square it. I've never done one like this. I was going out on a limb. But you are are correct because qroot7^4 is 49.

  16. marissalovescats
    • one year ago
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    So in that case it's -12 and not -6/5.

  17. primeralph
    • one year ago
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    @marissalovescats @Jhannybean Same thing.

  18. primeralph
    • one year ago
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    |dw:1373140157416:dw|

  19. marissalovescats
    • one year ago
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    Yes but the way she solved it was the correct way.

  20. primeralph
    • one year ago
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    |dw:1373140211718:dw|

  21. primeralph
    • one year ago
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    |dw:1373140263170:dw|

  22. marissalovescats
    • one year ago
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    I just put 7^2 because I knew you have to have like the same "base number" Or whatever.

  23. primeralph
    • one year ago
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    That 's if you want to use 7 and not sqrt(7)

  24. Jhannybean
    • one year ago
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    \[\large (\sqrt{7})^4 = 7^{4/2} = 7^2 \]

  25. Jhannybean
    • one year ago
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    Yeah....same thing.

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