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

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(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 $2^3 = 8$

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

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?

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