Question

How do you write $y={{10}^{x}}$ in log form?

Answer 1

We are given equation as $y={{10}^{x}}$ , we start solution by understanding the type of equation we have, then as we have to find the log form so we will learn how log and the given function connected to each other then we apply log on both side, we will use $\log {{a}^{b}}=b\log a$ to simplify, we will also use ${{\log }_{a}}a=1$ to get the most simplified term of the solution.

Complete step by step solution:
We are given an equation as $y={{10}^{x}}$ .
We can see that on the right side we have 10 which is raise to the power of x, means base is 10 and power is ‘x’, so our function is of the form $y={{a}^{b}}$ , so clearly it is an exponential function.
We are asked to convert it to the log form.
To do so we will start by understanding about log and exponential connection.
Log and exponential and the inverse function of one another, that is $\log {{\left( e \right)}^{x}}=x$ $\left( \log_{e} {e} \right)=x$ .
Now we have to change $y={{10}^{x}}$ in log.
We will apply log on both sides. As the base of the exponent is 10, so we will apply log with base 10.
We have $y={{10}^{x}}$ .
Applying log base 10, we get –
$\Rightarrow {{\log }_{10}}\left( y \right)={{\log }_{10}}\left( {{10}^{x}} \right)$ .
As we know that $\log \left( {{a}^{b}} \right)=b\log a$
So, $\log {{10}^{x}}=x\log 10$
So, our equation above become –
$\Rightarrow {{\log }_{10}}\left( y \right)=x{{\log }_{10}}\left( 10 \right)$
Now as we also know that ${{\log }_{a}}\left( a \right)=1$
So, ${{\log }_{10}}\left( 10 \right)=1$
So, using this in above statement, we get our equation ${{\log }_{10}}\left( y \right)=x{{\log }_{10}}\left( 10 \right)$ become ${{\log }_{10}}\left( y \right)=x$
So, our logarithmic term of the $y={{10}^{x}}$ becomes ${{\log }_{10}}\left( y \right)=x$.

Note: A short way to do so is to just use the relation between log and exponential.
When says that –
$a={{b}^{c}}\Leftrightarrow {{\log }_{b}}\left( a \right)=c$
So, $y={{10}^{x}}$ is same as ${{\log }_{10}}\left( y \right)=x$
We have other property of log as well like –
$\begin{aligned} & \log \left( xy \right)=\log x + \log y \\\ & \log \left( {}^{x}/{}_{y} \right)=\log x-\log y \\\ \end{aligned}$
Do not mistake by doing $\log \left( xy \right)=\log x+ \log y$ or $\log \left( {}^{x}/{}_{y} \right)=\dfrac{\log x}{\log y}$ .