If frac{log _{2} x}{3}=frac{log _{2} y}{4}=fr | vedclass

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(C) Given $\frac{\log _{2} x}{3}=\frac{\log _{2} y}{4}=\frac{\log _{2} z}{5 k} = \lambda$ (let).Then $\log _{2} x = 3\lambda$,$\log _{2} y = 4\lambda$

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$5$

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(C) Given $\frac{\log _{2} x}{3}=\frac{\log _{2} y}{4}=\frac{\log _{2} z}{5 k} = \lambda$ (let).Then $\log _{2} x = 3\lambda$,$\log _{2} y = 4\lambda$,and $\log _{2} z = 5k\lambda$.Given $\frac{z}{x^{3} y^{4}}=1$,taking $\log _{2}$ on both sides:$\log _{2} z - \log _{2} (x^{3} y^{4}) = \log _{2} 1$$\log _{2} z - 3 \log _{2} x - 4 \log _{2} y = 0$.Substituting the values of $\log _{2} x, \log _{2} y, \log _{2} z$:$5k\lambda - 3(3\lambda) - 4(4\lambda) = 0$$5k\lambda - 9\lambda - 16\lambda = 0$$5k\lambda - 25\lambda = 0$Since $\lambda 
eq 0$,we have $5k - 25 = 0$,which gives $k = 5$.

If $\frac{\log _{2} x}{3}=\frac{\log _{2} y}{4}=\frac{\log _{2} z}{5 k}$ and $\frac{z}{x^{3} y^{4}}=1,$ then $k=$

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