Question

How do you write $ {7^2} = 49 $ in log form?

Answer 1

Hint: In this question, we are given $ {7^2} = 49 $ . This form is known as exponential form. Exponent means how many times we are multiplying the same number to get the product. To convert an expression from exponential form we use the simple concept of conversion of exponential form to logarithmic form.

Complete step by step solution:
In this question, we are given
 $ {7^2} = 49 $ - - - - - - - (1)
And we are supposed to write it in logarithmic form.
The above given form is known as exponential form.
If an expression is given in exponential form,
 $ {a^m} = b $ - - - - - - - - - (2)
Then, in logarithmic form it can be written as
 $ {\log _a}b = m $ - - - - - - - - (3)
Here, it is called a base.
So, therefore from equation (2) and equation (3), we get
 $ {\log _7}49 = 2 $ as the logarithmic form of $ {7^2} = 49 $ .
Hence, we can write $ {7^2} = 49 $ as $ {\log _7}49 = 2 $ in logarithmic form.
We can also solve this with another method by introducing logs on both sides of the equation.
Introducing logs on both sides of equation (1), we get
 $ \Rightarrow \log {7^2} = \log 49 $
Now, we know that $ \log {a^b} = b\log a $
 $
   \Rightarrow 2\log 7 = \log 49 \\
   \Rightarrow \dfrac{{\log 49}}{{\log 7}} = 2 \;
  $
Now, we know that $ \dfrac{{\log a}}{{\log b}} = {\log _b}a $
 $ {\log _7}49 = 2 $
So, the correct answer is “ $ {\log _7}49 = 2 $ ”.

Note: We can never find the log of a negative value. It is only possible to find the logarithm of numbers greater than zero. Also the value of $ \log 0 $ is equal to 1 and the value of $ \log 1 $ is equal to 0.