Question

simplyfy [(3)^2]^3 + (2/3)^0 + 3^5 × (1/3)^4​

Answer 1

Given :- simplify [(3)^2]^3 + (2/3)^0 + 3^5 × (1/3)^4 ?

Solution :-

→ [(3)^2]^3 + (2/3)^0 + 3^5 × (1/3)^4

using :-

• (a^m)^n = a^(m * n)
• a^0 = 1

→ 3^(2 * 3) + 1 + 3^5 * (1/3⁴)

→ 3⁶ + 1 + 3⁵ * 1/3⁴

using :-

• 1/a^m = a^(-m)

→ 3⁶ + 1 + 3⁵ * 3^(-4)

using :-

• a^m * a^n = a^(m + n)

→ 3⁶ + 1 + 3^(5 - 4)

→ 729 + 1 + 3

→ 733 (Ans.)

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Answer 2

Given :  $[(3^2)]^3+(\frac{2}{3})^0+3^5\times(\frac{1}{3})^4$

To Find : Simplify.

Solution:

$[(3^2)]^3+(\frac{2}{3})^0+3^5\times(\frac{1}{3})^4$

Laws of exponents  

$\begin{align} & {{\text{a}}^{n}}\times {{a}^{-n}}=1\text{ or }{{\text{a}}^{n}}=\frac{1}{{{a}^{n}}} \\ & {{a}^{m}}\times {{a}^{n}}={{a}^{m+n}} \\ & \frac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}} \\ & {{\left( {{a}^{m}} \right)}^{n}}={{\left( {{a}^{n}} \right)}^{m}}={{a}^{mn}} \\ & {{a}^{m}}\times {{b}^{m}}={{\left( ab \right)}^{m}} \\ & \frac{{{a}^{n}}}{{{b}^{n}}}={{\left( \frac{a}{b} \right)}^{n}} \\ & {{a}^{0}}=1 \\ \end{align}$

$[(3^2)]^3+(\frac{2}{3})^0+3^5\times(\frac{1}{3})^4$

$=(3^6)+(\dfrac{2^0}{3^0})+3^5\times(\dfrac{1^4}{3^4})$

$=729+(\dfrac{1}{1})+ (\dfrac{3^5}{3^4})$

$=729+ 1 + 3^{5-4}$

$=729+ 1 + 3^1$

=  729+ 1 + 3

= 733

$[(3^2)]^3+(\frac{2}{3})^0+3^5\times(\frac{1}{3})^4=733$

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