Question #5119c

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Answer 1 · 2016-11-29T17:05:01

Let

$3^{x} = 7^{y} = 63^{z} = k$ $3^x=7^y=63^z=k$

So $3 = k^{\frac{1}{x}}$ $3=k^(1/x)$

$7 = k^{\frac{1}{y}}$ $7=k^(1/y)$

and

$63 = k^{\frac{1}{z}}$ $63=k^(1/z)$

$\Rightarrow 3^{2} \times 7 = k^{\frac{1}{z}}$ $=>3^2xx7=k^(1/z)$

$\Rightarrow {(k^{\frac{1}{x}})}^{2} \times k^{\frac{1}{y}} = k^{\frac{1}{z}}$ $=>(k^(1/x))^2xxk^(1/y)=k^(1/z)$

$\Rightarrow k^{\frac{2}{x}} \times k^{\frac{1}{y}} = k^{\frac{1}{z}}$ $=>k^(2/x)xxk^(1/y)=k^(1/z)$

$\Rightarrow k^{\frac{2}{x} + \frac{1}{y}} = k^{\frac{1}{z}}$ $=>k^(2/x+1/y)=k^(1/z)$

$\Rightarrow \frac{2}{x} + \frac{1}{y}) = \frac{1}{z}$ $=>2/x+1/y)=1/z$

$\Rightarrow \frac{2 y + x}{x y} = \frac{1}{z}$ $=>(2y+x)/(xy)=1/z$

$\Rightarrow z (2 y + x) = x y$ $=>z(2y+x)=xy$