Question

Draw the graph of $\log x$ .

Answer 1

Hint: For answering this question we need to draw a graph for the function $\log x$ . For that we will plot a curve for the respective values of $x$ and explore the basic definition of $\log x$ . The basic definition of $\log x$ says that for $a^y=x$ the logarithm value is given as $y={\log }_{a}x$ .

Complete step-by-step answer:
Now considering from the question we need to draw the graph for the function of $\log x$ .
From the basic definition of logarithm of a value for a function $a^y=x$ is given as $y={\log }_{a}x$ for $a>0,a\ne 1$ . Here $x>0$ is the domain for this function.
The graph will be continuous and smooth.
The curve is increasing for $a>1$ and decreasing for $0<a<1$ .
The curve of $\log x$ intersects the x-axis when $x=1$ .
The logarithm function is of one-one type.
The logarithm function is the inverse of the exponential function.
Properties:

(i) ${\log }_{a}1=0$ because $a^0=1$ .
(ii) ${\log }_{a}a=1$ because $a^1=a$ .
(iii) ${\log }_a{a}^x=x$ .
(iv) ${\log }_{a}x={\log }_{a}y\Rightarrow x=y$ .
(v) ${\log }_{a}x={\log }_{b}x\Rightarrow a=b$ .
Common logarithm has base 10 represented by $\log x$ .
Natural logarithm has a base $e$ represented by $\ln x$ .
Here we need to draw a graph of common logarithms.

Note: For answering questions based on logarithm functions we use different formulae like
$\log \left(ab\right)=\log a+\log b$, $\log \left({\displaystyle \frac{a}{b}}\right)=\log a-\log b$ and many more. The logarithm properties give that $\ln e=1$ and $\log 10=1$ . There is also one special property given as ${\log }_{a}b={\displaystyle \frac{\log b}{\log a}}$ and ${\log }_{a}b={\displaystyle \frac{1}{{\log }_{b}a}}$ . As this is an inverse function we can say that ${\log }_a\left(x^n\right)=n{\log }_{a}x$ and ${\log }_a\left(\sqrt[n]{x}\right)={\displaystyle \frac{1}{n}}{\log }_{a}x$ . We can say $a^{{\log }_{a}x}=x$ and ${\log }_{a}0=\left\{\begin{array}{c}-\mathrm{\infty }\text{ when a > 1}\\ \mathrm{\infty }\text{ when a < 1}\end{array}\right\}$ . We have another formulae saying that ${\log }_{a^m}{a}^n={\displaystyle \frac{n}{m}}$ for $m\ne 0$ .