Please solve sum no.25

D e a r s t u d e n t, 2^{x} = 3^{y} = 12^{z} L e t k = 2^{x} = 3^{y} = 12^{z} k = 2^{x} 2 = k^{\frac{1}{x}} s i m i l a r l y 3 = k^{\frac{1}{y}} a n d 12 = k^{\frac{1}{z}} N o w, 12 = 2 \times 2 \times 3 = 2^{2} \times 3 k^{\frac{1}{z}} = {(k^{\frac{1}{x}})}^{2} \times k^{\frac{1}{y}} k^{\frac{1}{z}} = k^{\frac{2}{x}} \times k^{\frac{1}{y}} k^{\frac{1}{z} - \frac{1}{y}} = k^{\frac{2}{x}} \Rightarrow \frac{1}{z} - \frac{1}{y} = \frac{2}{x} \frac{y - z}{z y} = \frac{2}{x} \frac{y - z}{2 z y} = \frac{1}{x} x = \frac{2 y z}{y - z} a n s R e g a r d s

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