यदि frac{log _{2} x}{3}=frac{log _{2} y}{4}=f | vedclass

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(C) माना कि $\frac{\log _{2} x}{3}=\frac{\log _{2} y}{4}=\frac{\log _{2} z}{5 k} = \lambda$.तब $\log _{2} x = 3\lambda$,$\log _{2} y = 4\lambda$,और $\

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$5$

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(C) माना कि $\frac{\log _{2} x}{3}=\frac{\log _{2} y}{4}=\frac{\log _{2} z}{5 k} = \lambda$.तब $\log _{2} x = 3\lambda$,$\log _{2} y = 4\lambda$,और $\log _{2} z = 5k\lambda$ प्राप्त होता है।दिया गया है $\frac{z}{x^{3} y^{4}}=1$,दोनों पक्षों में $\log _{2}$ लेने पर:$\log _{2} z - \log _{2} (x^{3} y^{4}) = \log _{2} 1$$\log _{2} z - 3 \log _{2} x - 4 \log _{2} y = 0$.$\log _{2} x, \log _{2} y, \log _{2} z$ के मान रखने पर:$5k\lambda - 3(3\lambda) - 4(4\lambda) = 0$$5k\lambda - 9\lambda - 16\lambda = 0$$5k\lambda - 25\lambda = 0$चूँकि $\lambda 
eq 0$,इसलिए $5k - 25 = 0$,जिससे $k = 5$ प्राप्त होता है।

यदि $\frac{\log _{2} x}{3}=\frac{\log _{2} y}{4}=\frac{\log _{2} z}{5 k}$ और $\frac{z}{x^{3} y^{4}}=1$ है,तो $k=$

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