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

1. 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

2. anonymous

$$\Large y=2^{\color{blue}{ x}}+5\impliedby hint$$

3. anonymous

Well, I already know what the equation is... I just don't understand how to solve :/

4. anonymous

we.. you asked to which it belongs to not to solve it

5. anonymous

6. anonymous

I guess I don't understand completely what the differences are.

7. anonymous

$$\Large y=(x+2)^{\frac{1}{2}+3}?$$

8. anonymous

well.. can't be... so I assume is hmmm $$\Large y=(x+2)^{\frac{1}{2}}+3?$$

9. anonymous

The +3 is not connected to the exponent

10. anonymous

Yes

11. anonymous

$$\bf y=(x+2)^{\frac{1}{2}}+3\implies y-3=(x+2)^{\frac{1}{2}} \\ \quad \\ y-3=\sqrt{x+2}\implies (y-3)^2=(\sqrt{x+2})^2\implies (y-3)^2=x+2 \\ \quad \\ (y-3)^2-2=x\implies y^2-6y+9-2=x \\ \quad \\ y^2-6y+7=x\impliedby quadratic$$

12. anonymous

Thank you so much! I will write this down in my notes!! :) Is the first one B?

13. anonymous

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

Oooooooh

15. anonymous

$$\Large { a^{\frac{{\color{blue} n}}{{\color{red} m}}} \implies \sqrt[{\color{red} m}]{a^{\color{blue} n}} \qquad \qquad \sqrt[{\color{red} m}]{a^{\color{blue} n}}\implies a^{\frac{{\color{blue} n}}{{\color{red} m}}} \\\quad \\ % rational negative exponent a^{-\frac{{\color{blue} n}}{{\color{red} m}}} = \cfrac{1}{a^{\frac{{\color{blue} n}}{{\color{red} m}}}} \implies \cfrac{1}{\sqrt[{\color{red} m}]{a^{\color{blue} n}}}\qquad\qquad % radical denominator \cfrac{1}{\sqrt[{\color{red} m}]{a^{\color{blue} n}}}= \cfrac{1}{a^{\frac{{\color{blue} n}}{{\color{red} m}}}}\implies a^{-\frac{{\color{blue} n}}{{\color{red} m}}} \\\quad \\ }$$ thus