How do you evaluate ${\log _{14}}\left| { - 1} \right|$ ?
Answer
579k+ views
Hint: In questions like this, we can assume the value which needs to be calculated as $x$ so that we can convert the equation from logarithmic form to exponential form and then we need to find a relation between left side of the equation and right side of the equation so that it makes a linear equation through which we can find the value of the variable $x$.
Complete step by step answer:
We are given to calculate the logarithmic expression ${\log _{14}}\left| { - 1} \right|$.
Let us assume ${\log _{14}}\left| { - 1} \right|$ as x.
So, $x = {\log _{14}}\left| { - 1} \right|$.
Now, we must evaluate the modulus function of the value inside the logarithm to get to the required answer. So, we know that the modulus function of a negative number yields the positive counterpart of the number. So, we get,
$ \Rightarrow x = {\log _{14}}\left( 1 \right)$
Since we know that if a and b are positive real numbers and b is not equal to $1$, then ${\log _b}a = y$ is equivalent to ${b^y} = a$.
Using the above-mentioned property here, we will get:
$ \Rightarrow {14^x} = 1$
Now, since we have to simplify the equation to solve it, we will try to make the base equal on both sides. So, as we know that ${14^0} = 1$, we will substitute $1$ as ${14^0}$ so as to make the bases equal on both sides of the equation.
$ \Rightarrow {14^x} = {14^0}$
Since the bases on both sides are the same, we can directly equate the exponential powers on them. So, now we can compare the powers of the variable on both sides of the equations.
We get, $x = 0$.
So, we get the value of x as $0$.
Note: The most important step here is the step where we have to figure out how to relate the left side of the equation and the right side of the equation so that it becomes easier to equate the exponents and obtain the value of variable x. In most of the questions, numbers are given in such a way that they can be expressed as the exponents of a same digit, so that the base of both sides of the equation becomes equal and we can easily equate their powers.
Complete step by step answer:
We are given to calculate the logarithmic expression ${\log _{14}}\left| { - 1} \right|$.
Let us assume ${\log _{14}}\left| { - 1} \right|$ as x.
So, $x = {\log _{14}}\left| { - 1} \right|$.
Now, we must evaluate the modulus function of the value inside the logarithm to get to the required answer. So, we know that the modulus function of a negative number yields the positive counterpart of the number. So, we get,
$ \Rightarrow x = {\log _{14}}\left( 1 \right)$
Since we know that if a and b are positive real numbers and b is not equal to $1$, then ${\log _b}a = y$ is equivalent to ${b^y} = a$.
Using the above-mentioned property here, we will get:
$ \Rightarrow {14^x} = 1$
Now, since we have to simplify the equation to solve it, we will try to make the base equal on both sides. So, as we know that ${14^0} = 1$, we will substitute $1$ as ${14^0}$ so as to make the bases equal on both sides of the equation.
$ \Rightarrow {14^x} = {14^0}$
Since the bases on both sides are the same, we can directly equate the exponential powers on them. So, now we can compare the powers of the variable on both sides of the equations.
We get, $x = 0$.
So, we get the value of x as $0$.
Note: The most important step here is the step where we have to figure out how to relate the left side of the equation and the right side of the equation so that it becomes easier to equate the exponents and obtain the value of variable x. In most of the questions, numbers are given in such a way that they can be expressed as the exponents of a same digit, so that the base of both sides of the equation becomes equal and we can easily equate their powers.
Recently Updated Pages
Write structures of the following compounds i 2 Chloro3methylpentane class 11 chemistry CBSE

What is BLO What is the full form of BLO class 8 social science CBSE

Explain the Treaty of Vienna of 1815 class 10 social science CBSE

A Paragraph on Pollution in about 100-150 Words

XIX+XXX A 49 B 51 C 55 D 44 class 5 maths CBSE

If x a + bt + ct2 where x is in meters and t is in class 11 physics CBSE

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Find the value of the expression given below sin 30circ class 11 maths CBSE

What do you mean by retardation What is its SI uni class 11 physics CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

10 examples of friction in our daily life

Difference between physical and chemical change class 11 chemistry CBSE

