Question

Given $x=lo{g}_{10}12$, $y=lo{g}_{4}2\times lo{g}_{10}9$ and $z=lo{g}_{10}0.4$, find :

(i) $x-y-z$

(ii) ${13}^{x-y-z}$

Answer 1

(i) x- y - z
= $lo{g}_{10}12-lo{g}_{4}2\times lo{g}_{10}9-lo{g}_{10}0.4$
= $lo{g}_{10}4\times 3-lo{g}_{4}2\times lo{g}_{10}9-lo{g}_{10}0.4$
=$lo{g}_{10}4\times 3-lo{g}_{4}2\times 2lo{g}_{10}3-lo{g}_{10}\frac{4}{10}$
= $lo{g}_{10}4+lo{g}_{10}3-\frac{lo{g}_2}{2lo{g}_2}\times 2lo{g}_{10}3-lo{g}_{10}4+lo{g}_{10}10$
= $lo{g}_{10}4+lo{g}_{10}3-\frac{2lo{g}_3}{2}-lo{g}_{10}4+1$
= 1

(ii)${13}^{x-y-z}$
= ${13}^1$
= 13 (since x-y-z = 1)