If frac{log _{2} a}{2}=frac{log _{3} b}{3}=fr | vedclass

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(D) Let $\frac{\log _{2} a}{2}=\frac{\log _{3} b}{3}=\frac{\log _{4} c}{4}=k$.From this,we get $\log _{2} a = 2k \Rightarrow a = 2^{2k}$,$\log _{3} b

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$c=256$

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(D) Let $\frac{\log _{2} a}{2}=\frac{\log _{3} b}{3}=\frac{\log _{4} c}{4}=k$.From this,we get $\log _{2} a = 2k \Rightarrow a = 2^{2k}$,$\log _{3} b = 3k \Rightarrow b = 3^{3k}$,and $\log _{4} c = 4k \Rightarrow c = 4^{4k}$.Substitute these into the given equation $a^{1/2} \cdot b^{1/3} \cdot c^{1/4} = 24$:$(2^{2k})^{1/2} \cdot (3^{3k})^{1/3} \cdot (4^{4k})^{1/4} = 24$$2^k \cdot 3^k \cdot 4^k = 24$$(2 \cdot 3 \cdot 4)^k = 24$$24^k = 24^1 \Rightarrow k = 1$.Now,substitute $k=1$ back into the expressions for $a, b,$ and $c$:$a = 2^{2(1)} = 4$$b = 3^{3(1)} = 27$$c = 4^{4(1)} = 256$.Thus,the correct option is $c=256$.

If $\frac{\log _{2} a}{2}=\frac{\log _{3} b}{3}=\frac{\log _{4} c}{4}$ and $a^{1 / 2} \cdot b^{1 / 3} \cdot c^{1 / 4}=24$,then:

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