Question

If $lo{g_4}7 = x$, then the value of $lo{g_7}16$ will be
A) ${x^2}$
B) $2x$
C) $x$
D) $\dfrac{2}{x}$

Answer 1

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	logb(x y) = y ∙ logb(x)
Logarithm of the base	logb(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.