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Question: How do you evaluate \(3{{\log }_{3}}9-4{{\log }_{3}}3\) ?...

How do you evaluate 3log⁡39−4log⁡333{{\log }_{3}}9-4{{\log }_{3}}3 ?

Explanation

Solution

If the value of ax{{a}^{x}} is equal to y then we can write x as log⁡ay{{\log }_{a}}y . Basically log⁡ay{{\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⁡ay{{\log }_{a}}y can be negative. Now we can easily find the value of 3log⁡39−4log⁡333{{\log }_{3}}9-4{{\log }_{3}}3 .

Complete step-by-step answer:
We have to evaluate 3log⁡39−4log⁡333{{\log }_{3}}9-4{{\log }_{3}}3
We know that ax{{a}^{x}} = y implies x = log⁡ay{{\log }_{a}}y
We know square of 3 is equal to 9 so log⁡39=2{{\log }_{3}}9=2 and we can write 3 as 3 to the power 1
So the value of log⁡33{{\log }_{3}}3 is equal to 1, replacing log⁡39{{\log }_{3}}9 with 2 and log⁡33{{\log }_{3}}3 with 1 in the equation we get
So we can write 3log⁡39−4log⁡33=3×2−4×13{{\log }_{3}}9-4{{\log }_{3}}3=3\times 2-4\times 1
Further solving we get the value of 3log⁡39−4log⁡333{{\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⁡ay{{\log }_{a}}y if a and y are both greater than 1 or both are less than 1 , the value of log⁡ay{{\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⁡ay{{\log }_{a}}y will be negative.