Question

Simplify:
(i) $(3^5{)}^{11}\times (3^{1}5{)}^4-(3^5{)}^{18}\times (3^5{)}^5$

(ii) $\frac{(16{)}^7\times (25{)}^5\times (81{)}^3}{(15{)}^7\times (24{)}^5\times (80{)}^3}$

Answer 1

We have
(i) $(3^5{)}^{11}\times (3^{15}{)}^4-(3^5{)}^{18}\times (3^5{)}^5$
$=3^{55}\times 3^{60}-3^{90}\times 3^{25}$ [$\because (a^m{)}^n=a^{m\times n}$]
$=3^{(55+60)}-3^{(90+25)}$ [$\because a^m\times a^n=a^{(m+n)}$]
$=3^{115}-3^{115}$
$=0$
(ii) $\frac{(16{)}^7\times (25{)}^5\times (81{)}^3}{(15{)}^7\times (24{)}^5\times (80{)}^3}$

$=\frac{(16{)}^7\times (5^2{)}^5\times (3^4{)}^3}{(3\times 5{)}^7\times (3\times 8{)}^5\times (16\times 5{)}^3}$

$=\frac{(16{)}^7\times (5{)}^{10}\times (3{)}^{12}}{3^7\times 5^7\times 3^5\times 8^5\times {16}^3\times 5^3}$ [$\because (a^m{)}^n=a^{m\times n},(a\times b{)}^n=a^n\times b^n$]

$=\frac{(16{)}^7\times (5{)}^{10}\times (3{)}^{12}}{3^7\times 3^5\times 5^7\times 5^3\times 8^5\times {16}^3}$

$=\frac{(16{)}^7\times (5{)}^{10}\times (3{)}^{12}}{3^{12}\times 5^{10}\times 8^5\times {16}^3}$ [$\because a^m\times a^n=a^{(m+n)}$]

$=\frac{(16{)}^7}{8^5\times {16}^3}$ [$\because a^m÷a^n=a^{(m-n)}$]

$=\frac{(16{)}^{7-3}}{8^5}$

$=\frac{(16{)}^4}{8^5}$

$=\frac{(2\times 8{)}^4}{8^5}$

$=\frac{2^4\times 8^4}{8^5}$

$=\frac{2^4}{8}=\frac{16}{8}=2$