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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

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"what type of function is that where x is a exponent" - they're called, somewhat not surprisingly, exponential functions... – J. M. Aug 3 '11 at 6:02
For what value of $x$ is $2^x$ the same as $1$, if you remember your laws of exponents? – J. M. Aug 3 '11 at 6:03

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 $0$.

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Of course, this assumes that the OP has a basic familiarity with logarithms... – Shai Covo Aug 3 '11 at 7:01

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,

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@MuffinMan: Regarding Patrick Da Silva's 2nd paragraph, to get a "right brain" explanation of what it means for a function to be injective (also called one-to-one), google the phrase "horizontal line test". Better yet, do an image search using both "horizontal line test" and "exponential" together. – Dave L. Renfro Aug 3 '11 at 13:54

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