How do you write \[{5^3} = 125\] in log form?

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

Complete step-by-step answer:
Given that,
 \[{5^3} = 125\]
This is of the form \[{b^y} = x\]
we know that \[\log {b^y} = y\log b\] so taking log on both sides
 \[ \Rightarrow 3\log 5 = \log 125\]
Taking logs on one side,
 \[ \Rightarrow 3 = \dfrac{{\log 125}}{{\log 5}}\]
We know that in logarithms we can write \[\dfrac{{\log x}}{{\log b}} = {\log _b}x\]
So applying this to above ratios also,
 \[ \Rightarrow 3 = {\log _5}125\]
Thus this is our final answer in the form of logarithm.
So, the correct answer is “ \[ 3 = {\log _5}125\] ”.

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 125 as a cube of 5 and then cancel both sides. That is not the way we have to solve this problem. We just have to write the logarithmic form of this exponential given.
And note one more thing exponential form and logarithmic form go hand in hand.