Why are there two different types of graph for logarithmic functions $\log_a X$ for different range of base,i.e; for : $0<a<1$ and $a>1$ ?
[Math] Why are there two different types of graph for logarithmic functions $\log_a{X}$ for different range of base,i.e., for : $01$
graphing-functionslogarithms
Related Solutions
To quote Wikipedia.
$\log_b a = {\log_d a \over \log_d b}$ This identity is useful to evaluate logarithms on calculators. For instance, most calculators have buttons for $\ln$ and for $\log_{10}$, but not for $\log_2$. To find $\log_2 3$, one could calculate $\frac{\log_{10} 3}{\log_{10} 2}$ (or $\frac{\ln 3}{\ln 2}$, which yields the same result).
So to calculate $\log_b a$ Just do $\frac{\ln a}{\ln b}$.
I suspect the reason they do this has to do with a Stack as the internal data structure of the calculator and Prefix / Postfix notation.
When you do $\ln a$ your first type $a$ and then $\ln$ So it primarily takes $1$ operand.
Too complex for a quiz.
Consider that you look for the zero's of function $$f(x)=a^x-\frac{\log (x)}{\log (a)}$$ Its derivative is given by $$f'(x)=a^x \log (a)-\frac{1}{x \log (a)}$$ this cancels at two points given by $$x_1=\frac{W_0\left(\frac{1}{\log (a)}\right)}{\log (a)}\qquad \text{and}\qquad x_2=\frac{W_{-1}\left(\frac{1}{\log (a)}\right)}{\log (a)}$$ where appears Lambert function. In the real domain, we need $a \lt e^{-e}$. When this is the case, $f(x_1)>0$ and $f(x_2) < 0$ and in this range $\lim_{x\to 0} \, f(x)=\infty$. So, for $0 < a < e^{-e}$, there are three roots (the first one between $0$ and $x_1$; the second one between $x_1$ and $x_2$; the third one above $x_2$); for $a=e^{-e}$, there is a triple root and for $a>e^{-e}$, there is a single root.
Edit
Since this is an interesting numerical problem, I give you below the three roots for a faw values of $a$ $$\left( \begin{array}{cccc} a & \text{first root} & \text{second root} & \text{third root} \\ 0.00500 & 0.005883 & 0.256675 & 0.969312 \\ 0.01000 & 0.013093 & 0.277987 & 0.941488 \\ 0.01500 & 0.021585 & 0.292615 & 0.913335 \\ 0.02000 & 0.031462 & 0.304205 & 0.884194 \\ 0.02500 & 0.042894 & 0.314008 & 0.853652 \\ 0.03000 & 0.056133 & 0.322619 & 0.821327 \\ 0.03500 & 0.071532 & 0.330371 & 0.786783 \\ 0.04000 & 0.089601 & 0.337471 & 0.749451 \\ 0.04500 & 0.111117 & 0.344056 & 0.708514 \\ 0.05000 & 0.137359 & 0.350225 & 0.662661 \\ 0.05500 & 0.170721 & 0.356048 & 0.609472 \\ 0.06000 & 0.216898 & 0.361580 & 0.543230 \\ 0.06500 & 0.303124 & 0.366862 & 0.436682 \\ 0.06510 & 0.306379 & 0.366965 & 0.433018 \\ 0.06520 & 0.309837 & 0.367069 & 0.429151 \\ 0.06530 & 0.313538 & 0.367172 & 0.425041 \\ 0.06540 & 0.317536 & 0.367275 & 0.420633 \\ 0.06550 & 0.321911 & 0.367378 & 0.415848 \\ 0.06560 & 0.326787 & 0.367481 & 0.410562 \\ 0.06570 & 0.332376 & 0.367584 & 0.404564 \\ 0.06580 & 0.339098 & 0.367686 & 0.397432 \\ 0.06590 & 0.348099 & 0.367789 & 0.388021 \\ 0.06591 & 0.349246 & 0.367799 & 0.386833 \\ 0.06592 & 0.350471 & 0.367810 & 0.385567 \\ 0.06593 & 0.351791 & 0.367820 & 0.384206 \\ 0.06594 & 0.353233 & 0.367830 & 0.382723 \\ 0.06595 & 0.354836 & 0.367840 & 0.381079 \\ 0.06596 & 0.356672 & 0.367851 & 0.379202 \\ 0.06597 & 0.358881 & 0.367861 & 0.376952 \\ 0.06598 & 0.361865 & 0.367871 & 0.373927 \end{array} \right)$$
For $a=e^{-e}$, the triple root is $0.367882$.
For the case of a single root $$\left( \begin{array}{cc} a & \text{ root} \\ 0.10 & 0.399013 \\ 0.15 & 0.436709 \\ 0.20 & 0.469622 \\ 0.25 & 0.500000 \\ 0.30 & 0.528956 \\ 0.35 & 0.557154 \\ 0.40 & 0.585043 \\ 0.45 & 0.612961 \\ 0.50 & 0.641186 \\ 0.55 & 0.669965 \\ 0.60 & 0.699535 \\ 0.65 & 0.730133 \\ 0.70 & 0.762013 \\ 0.75 & 0.795457 \\ 0.80 & 0.830785 \\ 0.85 & 0.868378 \\ 0.90 & 0.908699 \\ 0.95 & 0.952326 \end{array} \right)$$
Best Answer
From the definition of $\log x$ we can get, $\log_b{x}=\frac{\log{x}}{\log{b}}$. We are plotting $x$ in $x$-axis and $f(x)=\log_b{x}$ in $y$-axis, we will have $f(1)=0$.
Now, when we have $\boxed{\text{Case i:}}~~0<b<1$, for $0<x<1$ we will have $ \frac{\log{x}}{\log{b}}>0$, as $\log$ gives negative value for any input between $0$ and $1$. So, we will have positive value of $f(x)=\log_b{x}$, and for $x>1$ we have $\log{x}>0$ but, $\log{b}<0$, so,$f(x)=\frac{\log{x}}{\log{b}}<0$. Also, $\log{0}\to \infty$, as we go close to $0$, hence, the curve is not touching the $y$ -axis.
Another side, when we have $\boxed{\text{Case ii:}}$ $b>1$, $\log{b}>0$, now as we know $f(1)=\log_b{1}=0$, we need to check for $(i)~0<x<1$ and $(ii)~x>1$, to know the behavior of the graph. In this case, when $0<x<1$, $\log{x}<0$ making $f(x)= \frac{\log{x}}{\log{b}}<0$ and for $x>1$, we have $\log{x}>0$ making $f(x)= \frac{\log{x}}{\log{b}}>0$.