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**Hint:**We know that ${\log _e}a = k$and can also be written as $a = {e^k}$.

So we are going to use this formula to find the value of 4 from the given equation \[lo{g_4}7 = x\].

Then we substitute the value of ${4^2}$ in the equation \[lo{g_7}16\] for the value of 16 as $16 = {4^2}$ and find its value.

**Complete step-by-step answer:**

Given \[lo{g_4}7 = x\]……………………(1)

We have to find the value of \[lo{g_7}16\].

Now we know that ${\log _e}a = k$can also be written as $a = {e^k}$.

So by using the above equation we get from equation (1)

$ \Rightarrow lo{g_4}7 = x \Rightarrow 7 = {4^x}$

Now separating the 4 from the equation and taking x to the exponential of 7, we get

$ \Rightarrow 4 = {7^{\dfrac{1}{x}}}$

Now lets see the equation \[lo{g_7}16\]

Rules I used in solving the questions are mentioned below but we need to remember all the rules of logarithm to solve the questions correctly.

The base b logarithm of a number is the exponent that we need to raise the base in order to get the number.

Logarithm power rule | log_{b}(x ^{y}) = y ∙ log_{b}(x) |

Logarithm of the base | log_{b}(b) = 1 |

We know that ${\log _e}{x^y} = y.{\log _b}x$, then

\[lo{g_7}16 = lo{g_7}{4^2} = 2lo{g_7}4\]

Now substituting the value of 4 in the above equation from equation (2), we get

\[ = 2lo{g_7}\left( {{7^{\dfrac{1}{x}}}} \right)\]

We know that ${\log _e}{x^y} = y.{\log _b}x$, then

\[ = 2\left( {\dfrac{1}{x}} \right)lo{g_7}7\]

We know that ${\log _e}e = 1$, then

\[ = \dfrac{2}{x}\]

If \[lo{g_4}7 = x\], then the value of \[lo{g_7}16\] will be \[\dfrac{2}{x}\]

**So, option (D) is the correct answer.**

**Note:**Throughout your study of algebra, you have come across many properties—such as the commutative, associative, and distributive properties. These properties help you take a complicated expression or equation and simplify it. The same is true with logarithms. There are a number of properties that will help you simplify complex logarithmic expressions. So, one must know these properties to get the desired solution.

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