Question

How do you evaluate ${{\log }_{343}}49$?

Answer 1

Hint: We will solve the question by using the definition of logarithm in exponential form. We will use the property ${{\log }_{a}}x=b$ which is equal to ${{a}^{b}}=x$. So we will write the given expression in this form and simplify further to get the desired answer.

Complete step-by-step solution:
We know that the logarithm is the inverse function to the exponentiation. A logarithm is the power to which a number must be raised in order to get some other number.
We have been given that ${{\log }_{343}}49$.
We have to find the value of the given expression.
Now, we know that by definition of logarithm ${{\log }_{a}}x=b$ is equivalent to the ${{a}^{b}}=x$.
Now, let us assume that ${{\log }_{343}}49=b$.
So by using the above property we can write the expression as
\[\Rightarrow {{\left( 343 \right)}^{b}}=49\]
Now, we know that $49={{7}^{2}}$ and $343={{7}^{3}}$
Now, substituting the values in the above obtained equation we will get
\[\Rightarrow {{\left( {{7}^{3}} \right)}^{b}}={{7}^{2}}\]
Now, we know that \[{{\left( {{a}^{x}} \right)}^{b}}={{a}^{xb}}\]
So we will get
\[\Rightarrow \left( {{7}^{3b}} \right)={{7}^{2}}\]
Now, as bases are same so when we compare the power we will get
$\Rightarrow 3b=2$
Simplifying the above equation we will get
$\Rightarrow b=\dfrac{2}{3}$
So on solving the given expression we get the value ${{\log }_{343}}49=\dfrac{2}{3}$.

Note: To solve such types of questions students must have knowledge of logarithm and its properties. Logarithm rules are useful in expanding logarithms, condensing logarithms and solving logarithmic equations. As the given question is a simple question so be careful and avoid basic calculation mistakes.