I'm reviewing what I already learned in functions through my textbook except I can't get this question :
The question asks how to find the zeroes of $h(x) = 2^x -1$.
Also, I am wondering what type of function is that where $x$ is a exponent
I'm reviewing what I already learned in functions through my textbook except I can't get this question :
The question asks how to find the zeroes of $h(x) = 2^x -1$.
Also, I am wondering what type of function is that where $x$ is a exponent
It might also be useful for you to practice logarithms:
On the one hand, $ 2^x - 1 = 0 \Rightarrow 2^x = 1 \Rightarrow \ln (2^x ) = \ln (1) \Rightarrow x\ln 2 = 0 \Rightarrow x = 0. $ On the other hand, $ 2^0 - 1 = 1 - 1 = 0. $ Hence $ 2^x - 1 = 0 \Leftrightarrow x = 0. $ This means that $h$ has a unique zero at $0$.
Those functions (of the form $a^x$) are called exponentials. When you hear exponential in mathematical contexts though, it usually refers to $e^x$, where $e$ is Euler's constant.
Since $2^x$ is a strictly increasing function, it is injective, i.e. $h(x) = h(y)$ implies $x=y$. Therefore there is only one possible solution to your equation, if there is one. You can notice that $2^0 = 1$ and so $h(0) = 1-1 = 0$ and $0$ is the unique $0$ of $h$.
Hope that helps,