anonymous
  • anonymous
NEED HELP To which family does the function y=2^x+5 belong quadratic square root exponential logarithmic
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
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
jdoe0001
  • jdoe0001
\(\Large y=2^{\color{blue}{ x}}+5\impliedby hint\)
anonymous
  • anonymous
Well, I already know what the equation is... I just don't understand how to solve :/

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jdoe0001
  • jdoe0001
we.. you asked to which it belongs to not to solve it
jdoe0001
  • jdoe0001
seems solved already anyway
anonymous
  • anonymous
I guess I don't understand completely what the differences are.
jdoe0001
  • jdoe0001
\(\Large y=(x+2)^{\frac{1}{2}+3}?\)
jdoe0001
  • jdoe0001
well.. can't be... so I assume is hmmm \(\Large y=(x+2)^{\frac{1}{2}}+3?\)
anonymous
  • anonymous
The +3 is not connected to the exponent
anonymous
  • anonymous
Yes
jdoe0001
  • jdoe0001
\(\bf y=(x+2)^{\frac{1}{2}}+3\implies y-3=(x+2)^{\frac{1}{2}} \\ \quad \\ y-3=\sqrt{x+2}\implies (y-3)^2=(\sqrt{x+2})^2\implies (y-3)^2=x+2 \\ \quad \\ (y-3)^2-2=x\implies y^2-6y+9-2=x \\ \quad \\ y^2-6y+7=x\impliedby quadratic\)
anonymous
  • anonymous
Thank you so much! I will write this down in my notes!! :) Is the first one B?
jdoe0001
  • jdoe0001
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
anonymous
  • anonymous
Oooooooh
jdoe0001
  • jdoe0001
\(\Large { a^{\frac{{\color{blue} n}}{{\color{red} m}}} \implies \sqrt[{\color{red} m}]{a^{\color{blue} n}} \qquad \qquad \sqrt[{\color{red} m}]{a^{\color{blue} n}}\implies a^{\frac{{\color{blue} n}}{{\color{red} m}}} \\\quad \\ % rational negative exponent a^{-\frac{{\color{blue} n}}{{\color{red} m}}} = \cfrac{1}{a^{\frac{{\color{blue} n}}{{\color{red} m}}}} \implies \cfrac{1}{\sqrt[{\color{red} m}]{a^{\color{blue} n}}}\qquad\qquad % radical denominator \cfrac{1}{\sqrt[{\color{red} m}]{a^{\color{blue} n}}}= \cfrac{1}{a^{\frac{{\color{blue} n}}{{\color{red} m}}}}\implies a^{-\frac{{\color{blue} n}}{{\color{red} m}}} \\\quad \\ }\) thus

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