If $x^{18} = y^{21} = z^{28}$ , prove that $3$ , $3 {log}_{y} x, 3 {log}_{z} y,$ $7 {log}_{x} z$ form an A.P.

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Solution

Let $x^{18} = y^{21} = z^{28} = k$ , say then no taking log we get
$18$ log $x = 21$ log $y = 28$ log z $=$ log k .. $(1)$
If the given terms are a, b, c, d then
$a = 3, b = 3 l o g_{y} x = 3 (\frac{l o g x}{l o g y}) = 3 \cdot \frac{21}{18}$
$= \frac{21}{6} = \frac{7}{2} = 3 \frac{1}{2}$
Similarly, $c = 3 \cdot \frac{28}{21} = 4$ and $d = 7 \cdot \frac{18}{28} = \frac{9}{2} = 4 \frac{1}{2}$ .
Hence the four numbers are $3, 3 \frac{1}{2}, 4, 4 \frac{1}{2}$ which are clearly in A.P. of common difference $\frac{1}{2}$ .

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