Question

How do you evaluate $3{{\log }_{3}}9-4{{\log }_{3}}3$ ?

Answer 1

Hint: If the value of ${{a}^{x}}$ is equal to y then we can write x as ${{\log }_{a}}y$ . Basically ${{\log }_{a}}y$ represent the number of times we should multiply a to get y. In logarithm, the function base is always positive so the value of y has to be positive. ${{\log }_{a}}y$ can be negative. Now we can easily find the value of $3{{\log }_{3}}9-4{{\log }_{3}}3$ .

Complete step-by-step answer:
We have to evaluate $3{{\log }_{3}}9-4{{\log }_{3}}3$
We know that ${{a}^{x}}$ = y implies x = ${{\log }_{a}}y$
We know square of 3 is equal to 9 so ${{\log }_{3}}9=2$ and we can write 3 as 3 to the power 1
So the value of ${{\log }_{3}}3$ is equal to 1, replacing ${{\log }_{3}}9$ with 2 and ${{\log }_{3}}3$ with 1 in the equation we get
So we can write $3{{\log }_{3}}9-4{{\log }_{3}}3=3\times 2-4\times 1$
Further solving we get the value of $3{{\log }_{3}}9-4{{\log }_{3}}3$ is equal to 2.

Note: The base of logarithm in function is always taken positive because if we take base negative then the function will not be continuous it will be discrete because fraction power of negative number may not be a real number. But the value of logarithm can be negative in ${{\log }_{a}}y$ if a and y are both greater than 1 or both are less than 1 , the value of ${{\log }_{a}}y$ will be positive. If a is greater than 1 and y is less than 1 or a is less than 1 and y is greater than 1, ${{\log }_{a}}y$ will be negative.