The value of log_{20} 3 lies in which interva | vedclass

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

Question

(A) Let $x = \log_{20} 3$. This implies $20^x = 3$. We test the given intervals: If $x = \frac{1}{3}$,then $20^{1/3} = \sqrt[3]{20}$. Since $2^3 = 8$

Vedclass · Accepted Answer

$\left( \frac{1}{4}, \frac{1}{3} \right)$

Vedclass · Answer

(A) Let $x = \log_{20} 3$. This implies $20^x = 3$. We test the given intervals: If $x = \frac{1}{3}$,then $20^{1/3} = \sqrt[3]{20}$. Since $2^3 = 8$ and $3^3 = 27$,$\sqrt[3]{20}$ is between $2$ and $3$,so $20^{1/3} > 3$. Thus,$x If $x = \frac{1}{4}$,then $20^{1/4} = \sqrt[4]{20}$. Since $2^4 = 16$ and $3^4 = 81$,$\sqrt[4]{20}$ is slightly greater than $2$,so $20^{1/4}  \frac{1}{4}$. Therefore,$\frac{1}{4} < x < \frac{1}{3}$.

The value of $\log_{20} 3$ lies in which interval?

Similar Questions

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

If $a^x = b^y = c^z = d^w$,then the value of $x\left(\frac{1}{y} + \frac{1}{z} + \frac{1}{w}\right)$ is

The number of integers satisfying the inequality $\sqrt {{{\log }_3}(x) - 1} + \frac{{\frac{1}{2}{{\log }_3}({x^3})}}{{{{\log }_3}(\frac{1}{3})}} + 2 > 0$ is

If $y = \log_x x$, then the value of $\left( \frac{dy}{dx} \right)^2$ at $x = e^2$ is

If $\log _{1/\sqrt{2}} \sin x > 0$ for $x \in [0, 4\pi]$,find the number of values of $x$ that are integer multiples of $\frac{\pi}{4}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module