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anonymous

  • one year ago

NEED HELP To which family does the function y=2^x+5 belong quadratic square root exponential logarithmic

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  1. anonymous
    • one year ago
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    Same question for the equation y=(x+2)^(1/2)+3 The answers can be quadratic square root exponential reciprocal I know it's not exponential

  2. jdoe0001
    • one year ago
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    \(\Large y=2^{\color{blue}{ x}}+5\impliedby hint\)

  3. anonymous
    • one year ago
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    Well, I already know what the equation is... I just don't understand how to solve :/

  4. jdoe0001
    • one year ago
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    we.. you asked to which it belongs to not to solve it

  5. jdoe0001
    • one year ago
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    seems solved already anyway

  6. anonymous
    • one year ago
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    I guess I don't understand completely what the differences are.

  7. jdoe0001
    • one year ago
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    \(\Large y=(x+2)^{\frac{1}{2}+3}?\)

  8. jdoe0001
    • one year ago
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    well.. can't be... so I assume is hmmm \(\Large y=(x+2)^{\frac{1}{2}}+3?\)

  9. anonymous
    • one year ago
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    The +3 is not connected to the exponent

  10. anonymous
    • one year ago
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    Yes

  11. jdoe0001
    • one year ago
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    \(\bf y=(x+2)^{\frac{1}{2}}+3\implies y-3=(x+2)^{\frac{1}{2}} \\ \quad \\ y-3=\sqrt{x+2}\implies (y-3)^2=(\sqrt{x+2})^2\implies (y-3)^2=x+2 \\ \quad \\ (y-3)^2-2=x\implies y^2-6y+9-2=x \\ \quad \\ y^2-6y+7=x\impliedby quadratic\)

  12. anonymous
    • one year ago
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    Thank you so much! I will write this down in my notes!! :) Is the first one B?

  13. jdoe0001
    • one year ago
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    and yes, both have an exponent, and one could say they're both exponenttials BUT the difference being, one exponent is an rational or fraction the other is not so the rational can always be expressed as a root and from there you end up with some exponent on the other side

  14. anonymous
    • one year ago
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    Oooooooh

  15. jdoe0001
    • one year ago
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    \(\Large { a^{\frac{{\color{blue} n}}{{\color{red} m}}} \implies \sqrt[{\color{red} m}]{a^{\color{blue} n}} \qquad \qquad \sqrt[{\color{red} m}]{a^{\color{blue} n}}\implies a^{\frac{{\color{blue} n}}{{\color{red} m}}} \\\quad \\ % rational negative exponent a^{-\frac{{\color{blue} n}}{{\color{red} m}}} = \cfrac{1}{a^{\frac{{\color{blue} n}}{{\color{red} m}}}} \implies \cfrac{1}{\sqrt[{\color{red} m}]{a^{\color{blue} n}}}\qquad\qquad % radical denominator \cfrac{1}{\sqrt[{\color{red} m}]{a^{\color{blue} n}}}= \cfrac{1}{a^{\frac{{\color{blue} n}}{{\color{red} m}}}}\implies a^{-\frac{{\color{blue} n}}{{\color{red} m}}} \\\quad \\ }\) thus

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