Question

If $\frac{logx}{b-c}=\frac{logy}{c-a}=\frac{logz}{a-b}$, then what is the value of xyz?

Answer 1

$\frac{logx}{b-c}=\frac{logy}{c-a}=\frac{logz}{a-b}=k$

Therefore, $lo{g}_{10}x$ = k (b -c) ……. (i)

$lo{g}_{10}y$ = k (c -a) ……. (ii)

$lo{g}_{10}z$= k (a-b) ……. (iii)

Adding equations (i), (ii) and (iii) we get,

$lo{g}_{10}x$ + $lo{g}_{10}y$ + $lo{g}_{10}z$ = k (b -c) + k (c -a) + k (a-b)

$lo{g}_{10}xyz$ = k (b - c + c - a + a - b)

$lo{g}_{10}xyz$ = 0

xyz = ${10}^0$ (changing to exponential form)

xyz = 1