For a given number alpha 1,which of the fol | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) We know that for $\alpha > 1$,the value of $\log_{b}\alpha = \frac{\ln \alpha}{\ln b}$.Since $\ln \alpha$ is positive,the value of $\log_{b}\alpha

Vedclass · Accepted Answer

$\log_{10}\alpha, \log_{3}\alpha, \log_{e}\alpha, \log_{2}\alpha$

Vedclass · Answer

(B) We know that for $\alpha > 1$,the value of $\log_{b}\alpha = \frac{\ln \alpha}{\ln b}$.Since $\ln \alpha$ is positive,the value of $\log_{b}\alpha$ decreases as the base $b$ increases.The bases are $2, e \approx 2.718, 3, 10$.Since $2 < e < 3 < 10$,the values follow the order $\log_{10}\alpha < \log_{3}\alpha < \log_{e}\alpha < \log_{2}\alpha$.

For a given number $\alpha > 1$,which of the following is the correct ascending order?

Similar Questions

For the function $y = \log_a x$ to be defined,the base $a$ must be:

The solution of the equation $\log _{101} \log _{7}(\sqrt{x+7}+\sqrt{x})=0$ is

If $\log 2=a, \log 3=b, \log 7=c$ and $6^x=7^{x+4}$, then $x$ is equal to:

If ${x^{\frac{3}{4}(\log_3 x)^2 + \log_3 x - \frac{5}{4}} = \sqrt{3}}$,then $x$ is:

If $\sqrt{\log_3 x^{16}} + 9 \log_{27} \sqrt[3]{\frac{3}{x}} = 5$,then $x = \dots$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module