Question

NEED HELP

To which family does the function y=2^x+5 belong

quadratic
square root
exponential
logarithmic

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

seems solved already anyway

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

The +3 is not connected to the exponent

Answer 10

Yes

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

Oooooooh

Answer 15

\(\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