Question

simplify and express the result in power notation with positive exponent​

Answer 1

Solution:-1

$\\ \sf\longmapsto 7^5\times 7^3\times 7^{-4}$

$\\ \sf\longmapsto 7^{5+3+(-4)}$

$\\ \sf\longmapsto 7^{8-4}$

$\\ \sf\longmapsto 7^4$

Solution:-2

$\\ \sf\longmapsto 6^{11}\times 6^3\times 6^{-5}$

$\\ \sf\longmapsto 6^{11+3+(-5)}$

$\\ \sf\longmapsto 6^{14-5}$

$\\ \sf\longmapsto 6^9$

Solution:-3

$\\ \sf\longmapsto (-4)^{-2}\times 5^{-3}\times 5^{-4}$

$\\ \sf\longmapsto (\dfrac{1}{4})^2\times 5^{-3-4}$

$\\ \sf\longmapsto \dfrac{1}{16}\times 5^{-7}$

Solution:-4

$\\ \sf\longmapsto (\dfrac{5}{9})^{-2}\times (\dfrac{3}{5})^{-3}\times (\dfrac{3}{5})^{0}$

$\\ \sf\longmapsto (\dfrac{9}{5})^2\times (\dfrac{5}{3})^3\times 1$

$\\ \sf\longmapsto \dfrac{9^2}{5^2}\times \dfrac{5^3}{3^3}$

Answer 2

Answer:

a)

$\frac{ {7}^{5} \times {7}^{3} }{ {7}^{ - 4} } \\ = \frac{ {7}^{5 + 3} }{ {7}^{ - 4} } \\ = {7}^{5 + 3 - ( - 4)} \\ = {7}^{5 + 3 + 4} \\ = {7}^{12}$

b)

${6}^{11} \times {6}^{ 3} \times {6}^{ - 5} \\ = {6}^{11 + 3 + ( - 5)} \\ = {6}^{11 + 3 - 5} \\ = {6}^{9}$

c)

${( - 4)}^{ - 2} \times {5}^{ - 3} \times {5}^{ - 4} \\ = \frac{1}{ {( - 4)}^{2} } \times {5}^{ - 3 - 4} \\ = \frac{1}{16} \times {5}^{ - 7}\\=\:\frac{1}{16} \times\frac{1}{5^7}$

d)

${( \frac{5}{9} )}^{ - 2} \times {( \frac{3}{5} })^{ - 3} \times { (\frac{3}{5} })^{0} \\ = {( \frac{9}{5} )}^{2} \times {( \frac{5}{3} })^{3} \times 1 \\ = \frac{ {9}^{2} }{ {5}^{2} } \times \frac{ {5}^{3} }{ {3}^{3} } \\ = \frac{ ({3}^{2} )^{2} }{ {5}^{2} } \times \frac{ {5}^{3} }{ {3}^{3} } \\ = \frac{ {3}^{4} }{ {5}^{2} } \times \frac{ {5}^{3} }{ {3}^{3} } \\ = {3}^{4 - 3} \times {5}^{3 - 2} \\ = {3}^{1} \times {5}^{1} \\ = 3 \times 5 \\ = 15$

Note:

$> {a}^{m} \times {a}^{n} = {a}^{m + n} \\ \\ > \frac{ {a}^{m} }{ {a}^{n} } = {a}^{m - n}\\\\>\:a^0\:=\:1$