Question

How do you write $3 = {\log _2}8$ in exponential form?

Answer 1

Hint : Here we are given the numbers in logarithmic form. This logarithm is having base 2. But we are asked to find this form to be converted into exponential form. The given form is $y = {\log _b}x$ and we have to convert and find its exponential form as ${b^y} = x$ such that the power of that number is equal to the log of the answer written previously. So let’s start!

Complete step-by-step answer :
Given that $3 = {\log _2}8$
Given is logarithmic form.
Now we know that ${\log _b}x$ can be written as $\dfrac{{\log x}}{{\log b}}$ . So let’s write.
$\Rightarrow {\log _2}8 = \dfrac{{\log 8}}{{\log 2}}$
Substitute this in original log,
$\Rightarrow 3 = \dfrac{{\log 8}}{{\log 2}}$
Taking denominator on LHS we get,
$\Rightarrow 3\log 2 = \log 8$
We know that $y\log b = \log {b^y}$ thus applying the same here
$\Rightarrow \log {2^3} = \log 8$
Now either canceling the log or we can say taking antilog on both sides we write,
$\Rightarrow {2^3} = 8$
This is our exponential form of the logarithmic form so given.
So, the correct answer is “ ${2^3} = 8$ ”.

Note : Log and antilog are the exact opposite processes operated on a number .So they cancel each other. For finding these values we make use of log tables. But we can simply solve this as $y = {\log _b}x$ is written as ${b^y} = x$ . Note that natural logarithmic of x is generally written as $\ln x$ $\left( {{\text{is read as ln of x }}} \right)$ or ${\log _e}x$ $\left( {{\text{is read as log of x to the base e}}} \right)$ .
Students don’t write 8 as a cube of 2 and then cancel both logs. That is not the way we have to solve this problem. We just have to write the exponential form of this log given.