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anonymous

  • one year ago

Algebra 2 help would be greatly appreciated!

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  1. anonymous
    • one year ago
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    \[3^{2x}-15(3^{x})+56=0\]

  2. Astrophysics
    • one year ago
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    Hey so what's it asking

  3. Astrophysics
    • one year ago
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    Solve for x? :O

  4. anonymous
    • one year ago
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    Yeah solve for x

  5. madhu.mukherjee.946
    • one year ago
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    substitute 3^x as a and solve the equation as a quadratic equation

  6. madhu.mukherjee.946
    • one year ago
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    do you need more explanation ?i mean can do it on your own

  7. madhu.mukherjee.946
    • one year ago
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    @science00000 ??

  8. anonymous
    • one year ago
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    can i multiply the -15 and 3^x ???

  9. Astrophysics
    • one year ago
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    No but you can simplify it -5(3^x+1)

  10. Astrophysics
    • one year ago
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    \[-5(3^{x+1})\]

  11. anonymous
    • one year ago
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    don't seem to be quite understanding this :/

  12. Astrophysics
    • one year ago
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    You have to know your exponent rules, 5 x 3 = 15, so we have 5(3)(3^x) you can only do this with like bases so we have \[3^{x+1}\]

  13. Astrophysics
    • one year ago
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    So here is the general rule \[x^mx^n = x^{m+n}\]

  14. Astrophysics
    • one year ago
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    You can factor this I suppose

  15. Astrophysics
    • one year ago
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    Yes, factoring should work

  16. Astrophysics
    • one year ago
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    factoring you should end up with \[(3^x-8)(3^x-7)=0\] as we are treating 3^x as a, also suggested above

  17. anonymous
    • one year ago
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    ohh this makes a little more sense now!

  18. Astrophysics
    • one year ago
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    -7 x - 8 = 56 -7+-8=15

  19. ganeshie8
    • one year ago
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    \[\large 3^{2x}-15(3^{x})+56=0\] \[\large (\color{red}{3^{x}})^2-15(\color{red}{3^{x}})+56=0\] let \(t=\color{red}{3^{x}}\) \[\large (\color{red}t{})^2-15(\color{red}{t})+56=0\] you knw how to handle a quadratic

  20. Astrophysics
    • one year ago
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    -15*

  21. anonymous
    • one year ago
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    Yes!! Thanks everyone for helping! Really appreciate it.

  22. Astrophysics
    • one year ago
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    You can solve for x now, it involves logarithms, you know how to do that right? \[3^x - 7 = 0\] set it to 0 and solve for x for both

  23. anonymous
    • one year ago
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    Yeah of course. Thanks again everyone!

  24. Astrophysics
    • one year ago
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    Ok, yw :)

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