Question

Which is greater $ x={{\log }_{3}}5\text{ or y=lo}{{\text{g}}_{17}}25$ .

Answer 1

Hint: Start with the simplification of y using the identity $ {{\log }_{b}}{{a}^{c}}=c{{\log }_{b}}a={{\log }_{{{b}^{\dfrac{1}{c}}}}}a$ . Once you are done with the simplification, to compare between logarithmic functions, you need to ensure that they have the same base. So, use the identity $ {{\log }_{b}}a=\dfrac{1}{{{\log }_{a}}b}$ to get the logarithmic functions with the same base. Once you get two logarithmic functions with the same base, the function with higher value of x in the form log x is greater, provided base is greater than 1.

Complete step-by-step answer:
Let us start the simplification of y using the identity $ {{\log }_{b}}{{a}^{c}}=c{{\log }_{b}}a={{\log }_{{{b}^{\dfrac{1}{c}}}}}a$ .
$ y={{\log }_{17}}25={{\log }_{17}}{{5}^{2}}=2{{\log }_{17}}5={{\log }_{{{17}^{\dfrac{1}{2}}}}}5$
Now, we will use the identity $ {{\log }_{b}}a=\dfrac{1}{{{\log }_{a}}b}$ for both x and y to make the bases of the logarithmic part same, i.e., 5. On doing so, we get
 $ y={{\log }_{{{17}^{\dfrac{1}{2}}}}}5=\dfrac{1}{{{\log }_{5}}{{17}^{\dfrac{1}{2}}}}$
$ x={{\log }_{3}}5=\dfrac{1}{{{\log }_{5}}3}$
Now, if we compare $ {{\log }_{5}}{{17}^{\dfrac{1}{2}}}$ and $ {{\log }_{5}}3$ , we know , the function with higher value of x in the form log x is greater, provided base is greater than 1 and bases are same. So, $ {{17}^{\dfrac{1}{2}}}$ is for sure greater than 4 as the square of 4 is 16 which is less than 17. So, $ {{\log }_{5}}{{17}^{\dfrac{1}{2}}}$ is greater than $ {{\log }_{5}}3$ .
Now, as the two terms are the denominators of x and y, we can say that denominator of y is greater than x and both x and y have the same numerator. So, we can conclude that x>y.
Therefore, the answer to the above question is $ x>y$ .

Note: The most important thing to remember is that if two fractions have the same numerator and the denominator is different, then the fraction with the smaller denominator is greater, but the condition attached to this is both numerator and denominator needs to be positive and as the value of logarithmic function of a number is positive if the number and the base is greater than 1 which is true in the above solution.