Question

evaluate (64)⁷/⁶×(27)²/³×2-⁷÷(√81)³/².​

Answer 1

Answer:

$\frac{1}{3}$

Step-by-step explanation:

$= {(64)}^{7/6} \times {(27)}^{2/3} \times {2}^{ - 7} \div {( \sqrt{81} )}^{3/2} \\ = {( {2}^{6} )}^{7/6} \times {( {3}^{3} )}^{2/3} \times {2}^{ - 7} \div {( 9 )}^{3/2} \\ = {2}^{7} \times {3}^{2} \times {2}^{ - 7} \div { ({3}^{2} )}^{3/2} \\ = {3}^{2} \div {3}^{3} \\ = {3}^{2 - 3} \\ = {3}^{ - 1} \\ = \frac{1}{3}$

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Answer 2

Answer :-

• The answer is 1/3.

What to do?

• Evaluate (64)⁷/⁶ × (27)²/³ × 2-⁷ ÷ (√81)³/².

Laws of exponents to be used :-

$\boxed{\rm \bigstar \: {(a^{m} )}^{n} = {a}^{mn}}$

$\boxed{\rm\bigstar \: {a}^{m} \times {a}^{n} = {a}^{m + n}}$

$\boxed{\rm \bigstar \: {a}^{0} = 1}$

$\boxed{\rm \bigstar \: {a}^{m} \div {a}^{n} = {a}^{m - n}}$

$\boxed{\bigstar \: \rm {a}^{ - m} = \dfrac{1}{a^{m}}}$

Step-by-step explanation :-

• Here, we have to evaluate the expression which is given to us. As it contains powers, we will use some laws of exponents and powers to evaluate it!

$\longmapsto \sf(64)^{ \frac{7}{6} } \times (27)^{ \frac{2}{3} } \times 2^{ - 7} \div ( \sqrt{81} )^{ \frac{3}{2} }$

Expressing 64 as a power with base 2,

$\longmapsto\sf({2}^{6} )^{ \frac{7}{6} } \times (27)^{ \frac{2}{3} } \times 2^{ - 7} \div (\sqrt{81} )^{ \frac{3}{2} }$

Expressing 27 as a power with base 3,

$\longmapsto\sf ({2}^{6} )^{ \frac{7}{6} } \times ( {3}^{3} )^{ \frac{2}{3} } \times {2}^{ - 7} \div (\sqrt{81} )^{ \frac{3}{2} }$

Finding the square root of 81,

$\longmapsto\sf ({2}^{6} )^{ \frac{7}{6} } \times ( {3}^{3} )^{ \frac{2}{3} } \times {2}^{ - 7} \div (9)^{ \frac{3}{2} }$

Expressing 9 as a power with base 3,

$\longmapsto\sf ({2}^{6} )^{ \frac{7}{6} } \times ( {3}^{3} )^{ \frac{2}{3} } \times {2}^{ - 7} \div ( {3}^{2} )^{ \frac{3}{2} }$

As (a^m)^n = a^mn, therefore,

$\longmapsto\sf{2}^{ \not6\times\frac{7}{ \not6}} \times {3}^{ \not3 \times \frac{2}{ \not3} } \times {2}^{ - 7} \div {3}^{ \not2 \times \frac{3}{ \not2} }$

Cancelling the numbers,

$\longmapsto\sf{2}^{7} \times {3}^{2} \times {2}^{ - 7} \div {3}^{3}$

As a^m × a^n = a^m + n, therefore,

$\longmapsto \sf{2}^{7 + ( - 7)} \times {3}^{2} \div {3}^{3}$

$\longmapsto\sf {2}^{0} \times {3}^{2} \div {3}^{3}$

As a^m ÷ a^n = a^m - n, therefore,

$\longmapsto \sf \sf{2}^{0} \times {3}^{2 - 3}$

$\longmapsto\sf{2}^{0} \times {3}^{ - 1}$

As a^0 = 1, therefore,

$\longmapsto \sf1 \times {3}^{ - 1}$

As a^-m = 1/a^m, therefore,

$\longmapsto\sf1 \times \dfrac{1}{ {3}^{1} }$

$\longmapsto\sf1 \times \dfrac{1}{3}$

$\longmapsto \sf\dfrac{1}{3}$

Hence :-

• Our answer is 1/3.

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