How do you solve $ {{7}^{x}}=49 $ ?

Hint: The given question can be solved using the concept of the logarithm. In this question, we will apply log on both sides of the given equation and then proceed ahead with the basic logarithm operations in order to find the value of x.

Complete step by step answer:
The given equation is $ {{7}^{x}}=49 $ .
Applying $ \log $ on both the sides we get,
 $ \log {{7}^{x}}=\log 49 $
Using the property $ \log {{a}^{b}}=b\log a $ , we get
 $ \begin{align}
  & \Rightarrow x\log 7=\log 49 \\
 & \Rightarrow x\log 7=\log (7\times 7) \\
 & \Rightarrow x=\dfrac{\log (7\times 7)}{\log 7} \\
\end{align} $
Now we know that $ \log (a\times a)=\log a+\log a $ . Using this formula in the above equation we get,
 $ \Rightarrow x=\dfrac{\log 7+\log 7}{\log 7} $
Splitting it as two terms and simplifying further, we get
 $ \begin{align}
  & \Rightarrow x=\dfrac{\log 7}{\log 7}+\dfrac{\log 7}{\log 7} \\
 & \Rightarrow x=1+1 \\
 & \therefore x=2 \\
\end{align} $
Hence the value of $ x=2 $ .
Therefore, such type of questions can easily be solved by applying the concepts of logarithms and performing its basic operations.

Note:
Alternatively, the above question can also be solved by using the exponent’s concept
The given equation is $ {{7}^{x}}=49 $ …
Now the R.H.S $ 49 $ can be written as $ {{7}^{2}} $
Substituting $ 49 $ as $ {{7}^{2}} $ in the above equation we get,
 $ {{7}^{x}}={{7}^{2}} $
Now we know that when bases are same then the powers of RHS and LHS can be compared.
Hence on comparing the powers of the above equation we get $ x=2 $ .
 $ \therefore x=2 $
Students often do not perform the calculations of the $ \log $ carefully and solve the question wrong. Be aware of the common formulas of addition and subtraction used in the $ \log $. For better results learn all the formulas and the values of different $ \log $ numbers. Always remember to recheck the answer with the help of the alternate method of the exponents and powers as discussed above. $