Question

Evaluate the given expression ${{3}^{2-{{\log }_{3}}4}}$.

Answer 1

Hint: In this question, we will use the exponential formula which is given by ${{a}^{n-k}}={{a}^{n}}-{{a}^{k}}$ to evaluate the given expression. Then, we will find the value of ${{a}^{n}}$ and ${{a}^{k}}$ individually. Now, to solve the logarithmic function, we can use the formula, ${{\log }_{a}}x=\dfrac{{{\log }_{b}}(x)}{{{\log }_{b}}(a)}$. Then, we will solve it to get the result.

Complete step-by-step answer:
It is given in the question that we have to evaluate ${{3}^{2-{{\log }_{3}}4}}$. We know that ${{a}^{n-k}}$ can be written as ${{a}^{n}}-{{a}^{k}}$. So, we can write it as,
$={{3}^{2-{{\log }_{3}}4}}$
$={{3}^{2}}-{{3}^{{{\log }_{3}}4}}\ldots \ldots \ldots (i)$
Now, we will find the value of ${{\log }_{3}}4$ to expand this expression.
In order to calculate ${{\log }_{3}}4$ we will use the change of base rule. The rule is given by ${{\log }_{a}}x=\dfrac{{{\log }_{b}}(x)}{{{\log }_{b}}(a)}$. We can use this rule if (a) and (b) are greater than 0 and not equal to 1, and if x is greater than 0.
We can write ${{\log }_{10}}10$ as $\dfrac{\log 10}{\log 10}=1$, similarly we can write ${{\log }_{e}}10$ as $\dfrac{\log 10}{\log e}=2.302$. So, we can write ${{\log }_{3}}4$ as $\dfrac{\log 4}{\log 3}$. Here, the base can be 10 or $e$. We are considering the base as 10 in our solution, so we get,
$\Rightarrow \dfrac{{{\log }_{10}}4}{{{\log }_{10}}3}=\dfrac{0.602}{0.477}$
$=1.262$
On substituting the above value in equation (i), we will get,
$={{3}^{2}}-{{3}^{1.262}}\ldots \ldots \ldots (ii)$
As we know that ${{a}^{n}}-{{a}^{k}}={{a}^{n-k}}$, we can write equation (ii) as ${{3}^{2-1.262}}$, so we get
$={{3}^{2-1.262}}$
$={{3}^{0.738}}$
$=2.249\approx 2.25$
Therefore, the value of the given expression, ${{3}^{2-{{\log }_{3}}4}}$ is 2.25.

Note: While calculating ${{\log }_{3}}4$ we get $\dfrac{\log 4}{\log 3}$ by using base change rule and we have considered $\dfrac{{{\log }_{10}}4}{{{\log }_{10}}3}$ in our solution but we can also consider $\dfrac{{{\log }_{e}}4}{{{\log }_{e}}3}$. For both the cases, the final answer will be the same.