Let $f (x) = 2^x$. Show that... Let $f(x) = 2^x$. Show that $\dfrac{f(x+h) - f(x)}{h} = \dfrac{2^x(2^h-1)}{h}$.
First day of my precalc class in college, and I have no idea where to start to solve this one. Can anyone point me in the right direction?
 A: Start by plugging in what you know. You're trying to figure out what $(f(x+h) - f(x))/h$ is equal to, and you know what $f(x)$ equals. So why don't you plug in all that and see what you get. From there, just remember the exponent formula $x^{a +b} = (x^a)(x^b)$.
A: $\frac{f(x+h)-f(x)}{h}=\frac{2^{(x+h)}-2^x}{h}$using laws of exponentiation we know that for real numbers a,b and c $a^b*a^c=a^{(b+c)}$using this we get that $\frac{2^{(x+h)}-2^x}{h}=\frac{2^x(2^h-1)}{h}  $
A: Let $f(x)=2^x$, then $f(x+h)$ is the same that replace the original function by $x+h$, then $f(x+h)=2^{(x+h)}$, which by basic school algebra is $2^x$ times $2^h$, then, you can replace your function $\dfrac{f(x+h) + f(x)}{h}$ by $\dfrac{(2^x)(2^h)+2^x}{h}$ if you factorize it, you get $\dfrac{2^x(2^h+1)}{h}$ which is what you nedded, then you get your answer. 
A: You have $f(x) = 2^x$. So for example you would have $f(2) = 2^2$ and $f(7) = 2^7$. So, when you evaluate $f$ at some number, then you just put that number in where the $x$ is.
You also have for example $f(x + h) = 2^{x+ h}$. So:
$$\dfrac{f(x+h) - f(x)}{h} = \dfrac{2^{x + h} - 2^x}{h}.$$
All you do then is factor out an $2^x$ in the numerator (which I will assume that you know how to do).
