Hello, I've trouble solving "graph log2x<0" how do i go about doing this? I know that log2x=y is equivalent to 2^y=x. However, the inequality sign and the <0 makes it complicated. The given answer is 1>x>0.
$\displaystyle \log_2(x) = \dfrac{\ln x}{\ln 2}$
If $\displaystyle y>0$, then $\displaystyle 2^{-y} = \dfrac{1}{2^y}$. If you increase $\displaystyle y$, $\displaystyle \dfrac{1}{2^y}$ gets "closer" to zero. You should know that there is no value for $\displaystyle y$ that makes the equation $\displaystyle 2^y = 0$ true, so $\displaystyle \log_2 (0)$ is not defined. Thus, $\displaystyle x>0$. Instead of increasing $\displaystyle y$, let's decrease it. As $\displaystyle y$ gets closer to zero, $\displaystyle 2^{-y}$ gets closer to 1, but is always less than one. So, that tells you $\displaystyle x<1$.
You graph this inequality the exact same way you graph ANY inequality.
First, you graph the related equality. In this case that seems to be $y = log_2(x).$ Is that function defined for non-positive values of x?
Second, you identify on what side of the point or points of equality, the inequality is true.
So at what value of x does $log_2(x) = 0.$
To the left of that point what is the sign of the function? To the right of that point what is the sign of the function?
Well the domain of $\displaystyle \begin{align*} \log_2{(x)} \end{align*}$ is $\displaystyle \begin{align*} x > 0 \end{align*}$. Solving the inequality for x gives
$\displaystyle \begin{align*} \log_2{(x)} &< 0 \\ x &< 2^0 \\ x &< 1 \end{align*}$
Thus, when combined with the domain of this function, the solution must be $\displaystyle \begin{align*} 0 < x < 1 \end{align*}$.
This has only the single variable x, rather than, say x and y. Its graph will be a portion of a number line, not a two dimensional x, y graph. In general, the regions where f(x)< a and f(x)> a are separated by points where f(x)= a. Here, mark the single point x such that $\displaystyle log_2(x)= 1$. Then, perhaps, choose one value of x less than that and one value of x larger to determine which side satisfies $\displaystyle log_2(x)< 1$ and which satisfies $\displaystyle k=log_2(x)> 1$.