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jonzion1122

  • 2 years ago

3^X+3 - 9^2X-1=0 what are the steps to finding the solution?

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  1. myininaya
    • 2 years ago
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    \[3^x+3-9^{2x}-1=0?\]

  2. myininaya
    • 2 years ago
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    Or! Do you mean \[3^{x+3}-9^{2x-1}=0\]

  3. jonzion1122
    • 2 years ago
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    second one

  4. myininaya
    • 2 years ago
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    Ok did you add to both sides \[9^{2x+1}\]

  5. myininaya
    • 2 years ago
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    \[3^{x+2}=9^{2x+1}\]

  6. myininaya
    • 2 years ago
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    oops - not + oh well

  7. myininaya
    • 2 years ago
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    Do you know how to answer the following What number satisfies the equation \[9=3^{?}\]

  8. myininaya
    • 2 years ago
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    Ok great. So we have \[3^{x+2}=(3^{2})^{(2x-1)}\] since 3^2=9 :)

  9. myininaya
    • 2 years ago
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    So we have the same base on both sides All you need to do is set the exponents equal and solve for x

  10. myininaya
    • 2 years ago
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    \[x+2=2(2x-1)\]

  11. myininaya
    • 2 years ago
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    Let me know if you have further questions :)

  12. jonzion1122
    • 2 years ago
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    firstly its x+3 but where did you get the 2 in "2"(2x-1)

  13. myininaya
    • 2 years ago
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    \[9=3^2 \] \[3^{x+3}=9^{2x-1}\] \[3^{x+2}=(3^2)^{(2x-1)}\]

  14. jonzion1122
    • 2 years ago
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    how does 3^2(^2x-1) = 2(2x-1)

  15. myininaya
    • 2 years ago
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    No....we have x+2=2(2x-1) not what you wrote

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