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Answer 1

Question: If $\sf a^{x} = b^{y}=c^{z}$ and b² = ac, prove that $\sf y = \dfrac{2xz}{x + z}$

Solution:

$\large{\underline{\underline{\sf \star \: Consider \: that :}}}$

$\sf a^{x} = b^{y}=c^{z} = k \: \: (suppose)$

Therefore,

$\sf a^{x} = k \\ \implies \sf a = k {}^{ \frac{1}{x} } \: \: ....(i) \\ \\ \sf b^{y}= k\\ \implies \sf b = k ^{\frac{1}{y}} \: \: (ii) \\ \\ \sf c^{z} = k \\ \implies \sf c = k {}^{ \frac{1}{z} } \: \: (iii)$

Now, take b² =ac and put the above values -

$\implies \sf b {}^{2} = ac \\ \\ \implies \sf (k {}^{ \frac{1}{y} } ) {}^{2} = (k {}^{ \frac{1}{x} } ) \times (k {}^{ \frac{1}{z} } ) \\ \\ \implies \sf (k ){}^{ \frac{2}{y} } = (k) {}^{ \frac{1}{x} + \frac{1}{z} } \\ \\ \bf on \: comparing \: both \: sides - \\ \\ \implies \sf \frac{2}{y} = \frac{1}{x} + \frac{1}{z} \\ \\ \implies \sf \frac{2}{y} = \frac{x + z}{xz} \\ \\ \bf on \: cross \: multiplying - \\ \\ \implies \sf y(x + z) = 2xz \\ \\ \boxed{ \bf y = \frac{2xz}{x + z}}$

Hence, it is proved.