If log _{2} x+log _{4} x+log _{16} x=21 / 4, | vedclass

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(A) Given the equation: $\log _{2} x+\log _{4} x+\log _{16} x=\frac{21}{4}$Using the change of base formula $\log _{a^n} x = \frac{1}{n} \log _{a} x$,

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$8$

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(A) Given the equation: $\log _{2} x+\log _{4} x+\log _{16} x=\frac{21}{4}$Using the change of base formula $\log _{a^n} x = \frac{1}{n} \log _{a} x$,we can rewrite the terms:$\log _{4} x = \log _{2^2} x = \frac{1}{2} \log _{2} x$$\log _{16} x = \log _{2^4} x = \frac{1}{4} \log _{2} x$Substituting these into the original equation:$\log _{2} x + \frac{1}{2} \log _{2} x + \frac{1}{4} \log _{2} x = \frac{21}{4}$Factor out $\log _{2} x$:$\log _{2} x \left(1 + \frac{1}{2} + \frac{1}{4}ight) = \frac{21}{4}$$\log _{2} x \left(\frac{4+2+1}{4}ight) = \frac{21}{4}$$\log _{2} x \left(\frac{7}{4}ight) = \frac{21}{4}$$\log _{2} x = \frac{21}{4} 	imes \frac{4}{7} = 3$Converting from logarithmic to exponential form:$x = 2^3 = 8$

If $\log _{2} x+\log _{4} x+\log _{16} x=21 / 4,$ then $x=$

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