Answer :
Sure, let's go through these step-by-step:
### a) [tex]\(\left(2^2\right)^3\)[/tex]
First, evaluate the exponent inside the parentheses:
[tex]\[ 2^2 = 4 \][/tex]
Next, raise that result to the power of 3:
[tex]\[ 4^3 = 64 \][/tex]
So, [tex]\(\left(2^2\right)^3 = 64 \)[/tex].
### b) [tex]\(\left(3^6\right)^{\frac{1}{2}}\)[/tex]
First, evaluate the exponent inside the parentheses:
[tex]\[ 3^6 = 729 \][/tex]
Next, take the square root of 729 (as [tex]\(\frac{1}{2}\)[/tex] represents the square root):
[tex]\[ \sqrt{729} = 27 \][/tex]
So, [tex]\(\left(3^6\right)^{\frac{1}{2}} = 27\)[/tex].
### c) [tex]\(\left(5^2\right)^{\frac{1}{2}}\)[/tex]
First, evaluate the exponent inside the parentheses:
[tex]\[ 5^2 = 25 \][/tex]
Next, take the square root of 25:
[tex]\[ \sqrt{25} = 5 \][/tex]
So, [tex]\(\left(5^2\right)^{\frac{1}{2}} = 5\)[/tex].
### d) [tex]\(\left(7^{-3}\right)^{\frac{1}{3}}\)[/tex]
First, evaluate the exponent inside the parentheses:
[tex]\[ 7^{-3} = \frac{1}{7^3} = \frac{1}{343} \][/tex]
Next, take the cube root of [tex]\(\frac{1}{343}\)[/tex] (as [tex]\(\frac{1}{3}\)[/tex] represents the cube root):
[tex]\[ \left(\frac{1}{343}\right)^{\frac{1}{3}} = \frac{1}{7} \approx 0.142857 \][/tex]
So, [tex]\(\left(7^{-3}\right)^{\frac{1}{3}} = 0.142857\)[/tex].
### f) [tex]\(4^{\frac{1}{2}}\)[/tex]
This expression represents the square root of 4:
[tex]\[ \sqrt{4} = 2 \][/tex]
So, [tex]\(4^{\frac{1}{2}} = 2\)[/tex].
### g) [tex]\(9^{\frac{1}{2}}\)[/tex]
This expression represents the square root of 9:
[tex]\[ \sqrt{9} = 3 \][/tex]
So, [tex]\(9^{\frac{1}{2}} = 3\)[/tex].
### h) [tex]\(8^{\frac{2}{3}}\)[/tex]
Express 8 as [tex]\(2^3\)[/tex]:
[tex]\[ 8 = 2^3 \][/tex]
Next, raise [tex]\(2^3\)[/tex] to the [tex]\(\frac{2}{3}\)[/tex] power:
[tex]\[ \left(2^3\right)^{\frac{2}{3}} = 2^{(3 \cdot \frac{2}{3})} = 2^2 = 4 \][/tex]
So, [tex]\(8^{\frac{2}{3}} = 4\)[/tex].
### i) [tex]\(16^{\frac{1}{4}}\)[/tex]
Express 16 as [tex]\(2^4\)[/tex]:
[tex]\[ 16 = 2^4 \][/tex]
Next, take the fourth root of 16:
[tex]\[ \left(2^4\right)^{\frac{1}{4}} = 2^{(4 \cdot \frac{1}{4})} = 2^1 = 2 \][/tex]
So, [tex]\(16^{\frac{1}{4}} = 2\)[/tex].
### e) [tex]\(\left(\frac{2^2}{3^2}\right)^{-\frac{1}{2}}\)[/tex]
First, calculate the initial expression:
[tex]\[ \frac{2^2}{3^2} = \frac{4}{9} \][/tex]
Next, take the square root of [tex]\(\frac{4}{9}\)[/tex] and then take the reciprocal due to the negative exponent:
[tex]\[ \left(\frac{4}{9}\right)^{-\frac{1}{2}} = \frac{1}{\sqrt{\frac{4}{9}}} = \frac{1}{\frac{2}{3}} = \frac{3}{2} = 1.5 \][/tex]
So, [tex]\(\left(\frac{2^2}{3^2}\right)^{-\frac{1}{2}} = 1.5\)[/tex].
### k) [tex]\(81^{-\frac{3}{4}}\)[/tex]
Express 81 as [tex]\(3^4\)[/tex]:
[tex]\[ 81 = 3^4 \][/tex]
Next, take the fourth root and then raise it to the negative third power:
[tex]\[ \left(3^4\right)^{-\frac{3}{4}} = 3^{(4 \cdot -\frac{3}{4})} = 3^{-3} = \frac{1}{3^3} = \frac{1}{27} \approx 0.037037 \][/tex]
So, [tex]\(81^{-\frac{3}{4}} = 0.037037\)[/tex].
### l) [tex]\(\left(\frac{16}{25}\right)^{\frac{1}{2}}\)[/tex]
First, take the square root of [tex]\(\frac{16}{25}\)[/tex]:
[tex]\[ \left(\frac{16}{25}\right)^{\frac{1}{2}} = \sqrt{\frac{16}{25}} = \frac{\sqrt{16}}{\sqrt{25}} = \frac{4}{5} = 0.8 \][/tex]
So, [tex]\(\left(\frac{16}{25}\right)^{\frac{1}{2}} = 0.8\)[/tex].
### m) [tex]\(\left(\frac{16}{81}\right)^{-\frac{1}{4}}\)[/tex]
First, take the fourth root of [tex]\(\frac{16}{81}\)[/tex] and then take the reciprocal due to the negative exponent:
[tex]\[ \left(\frac{16}{81}\right)^{-\frac{1}{4}} = \frac{1}{\left(\frac{16}{81}\right)^{\frac{1}{4}}} = \frac{1}{\frac{2}{3}} = \frac{3}{2} \][/tex]
So, [tex]\(\left(\frac{16}{81}\right)^{-\frac{1}{4}} = 1.5\)[/tex].
### n) [tex]\(\left(\frac{625}{81}\right)^{-0.5}\)[/tex]
First, take the square root of [tex]\(\frac{625}{81}\)[/tex] and then take the reciprocal due to the negative exponent:
[tex]\[ \left(\frac{625}{81}\right)^{-0.5} = \frac{1}{\left(\frac{625}{81}\right)^{0.5}} = \frac{1}{\frac{25}{9}} = \frac{9}{25} = 0.36 \][/tex]
So, [tex]\(\left(\frac{625}{81}\right)^{-0.5} = 0.36\)[/tex].
### j) [tex]\(8^{-\frac{1}{3}}\)[/tex]
Express 8 as [tex]\(2^3\)[/tex]:
[tex]\[ 8 = 2^3 \][/tex]
Next, take the cube root and then take the reciprocal due to the negative exponent:
[tex]\[ \left(2^3\right)^{-\frac{1}{3}} = 2^{3 \cdot -\frac{1}{3}} = 2^{-1} = \frac{1}{2} \][/tex]
So, [tex]\(8^{-\frac{1}{3}} = 0.5\)[/tex].
### o) [tex]\(\left(\frac{243}{32}\right)^{-0.4}\)[/tex]
Take the expression to the power of [tex]\(-0.4\)[/tex]:
[tex]\[ \left(\frac{243}{32}\right)^{-0.4} = \left(\frac{32}{243}\right)^{0.4} \][/tex]
Although this result was directly computed, its numerical value is approximately:
[tex]\[ 0.444444 \][/tex]
So, [tex]\(\left(\frac{243}{32}\right)^{-0.4} = 0.444444\)[/tex].
### a) [tex]\(\left(2^2\right)^3\)[/tex]
First, evaluate the exponent inside the parentheses:
[tex]\[ 2^2 = 4 \][/tex]
Next, raise that result to the power of 3:
[tex]\[ 4^3 = 64 \][/tex]
So, [tex]\(\left(2^2\right)^3 = 64 \)[/tex].
### b) [tex]\(\left(3^6\right)^{\frac{1}{2}}\)[/tex]
First, evaluate the exponent inside the parentheses:
[tex]\[ 3^6 = 729 \][/tex]
Next, take the square root of 729 (as [tex]\(\frac{1}{2}\)[/tex] represents the square root):
[tex]\[ \sqrt{729} = 27 \][/tex]
So, [tex]\(\left(3^6\right)^{\frac{1}{2}} = 27\)[/tex].
### c) [tex]\(\left(5^2\right)^{\frac{1}{2}}\)[/tex]
First, evaluate the exponent inside the parentheses:
[tex]\[ 5^2 = 25 \][/tex]
Next, take the square root of 25:
[tex]\[ \sqrt{25} = 5 \][/tex]
So, [tex]\(\left(5^2\right)^{\frac{1}{2}} = 5\)[/tex].
### d) [tex]\(\left(7^{-3}\right)^{\frac{1}{3}}\)[/tex]
First, evaluate the exponent inside the parentheses:
[tex]\[ 7^{-3} = \frac{1}{7^3} = \frac{1}{343} \][/tex]
Next, take the cube root of [tex]\(\frac{1}{343}\)[/tex] (as [tex]\(\frac{1}{3}\)[/tex] represents the cube root):
[tex]\[ \left(\frac{1}{343}\right)^{\frac{1}{3}} = \frac{1}{7} \approx 0.142857 \][/tex]
So, [tex]\(\left(7^{-3}\right)^{\frac{1}{3}} = 0.142857\)[/tex].
### f) [tex]\(4^{\frac{1}{2}}\)[/tex]
This expression represents the square root of 4:
[tex]\[ \sqrt{4} = 2 \][/tex]
So, [tex]\(4^{\frac{1}{2}} = 2\)[/tex].
### g) [tex]\(9^{\frac{1}{2}}\)[/tex]
This expression represents the square root of 9:
[tex]\[ \sqrt{9} = 3 \][/tex]
So, [tex]\(9^{\frac{1}{2}} = 3\)[/tex].
### h) [tex]\(8^{\frac{2}{3}}\)[/tex]
Express 8 as [tex]\(2^3\)[/tex]:
[tex]\[ 8 = 2^3 \][/tex]
Next, raise [tex]\(2^3\)[/tex] to the [tex]\(\frac{2}{3}\)[/tex] power:
[tex]\[ \left(2^3\right)^{\frac{2}{3}} = 2^{(3 \cdot \frac{2}{3})} = 2^2 = 4 \][/tex]
So, [tex]\(8^{\frac{2}{3}} = 4\)[/tex].
### i) [tex]\(16^{\frac{1}{4}}\)[/tex]
Express 16 as [tex]\(2^4\)[/tex]:
[tex]\[ 16 = 2^4 \][/tex]
Next, take the fourth root of 16:
[tex]\[ \left(2^4\right)^{\frac{1}{4}} = 2^{(4 \cdot \frac{1}{4})} = 2^1 = 2 \][/tex]
So, [tex]\(16^{\frac{1}{4}} = 2\)[/tex].
### e) [tex]\(\left(\frac{2^2}{3^2}\right)^{-\frac{1}{2}}\)[/tex]
First, calculate the initial expression:
[tex]\[ \frac{2^2}{3^2} = \frac{4}{9} \][/tex]
Next, take the square root of [tex]\(\frac{4}{9}\)[/tex] and then take the reciprocal due to the negative exponent:
[tex]\[ \left(\frac{4}{9}\right)^{-\frac{1}{2}} = \frac{1}{\sqrt{\frac{4}{9}}} = \frac{1}{\frac{2}{3}} = \frac{3}{2} = 1.5 \][/tex]
So, [tex]\(\left(\frac{2^2}{3^2}\right)^{-\frac{1}{2}} = 1.5\)[/tex].
### k) [tex]\(81^{-\frac{3}{4}}\)[/tex]
Express 81 as [tex]\(3^4\)[/tex]:
[tex]\[ 81 = 3^4 \][/tex]
Next, take the fourth root and then raise it to the negative third power:
[tex]\[ \left(3^4\right)^{-\frac{3}{4}} = 3^{(4 \cdot -\frac{3}{4})} = 3^{-3} = \frac{1}{3^3} = \frac{1}{27} \approx 0.037037 \][/tex]
So, [tex]\(81^{-\frac{3}{4}} = 0.037037\)[/tex].
### l) [tex]\(\left(\frac{16}{25}\right)^{\frac{1}{2}}\)[/tex]
First, take the square root of [tex]\(\frac{16}{25}\)[/tex]:
[tex]\[ \left(\frac{16}{25}\right)^{\frac{1}{2}} = \sqrt{\frac{16}{25}} = \frac{\sqrt{16}}{\sqrt{25}} = \frac{4}{5} = 0.8 \][/tex]
So, [tex]\(\left(\frac{16}{25}\right)^{\frac{1}{2}} = 0.8\)[/tex].
### m) [tex]\(\left(\frac{16}{81}\right)^{-\frac{1}{4}}\)[/tex]
First, take the fourth root of [tex]\(\frac{16}{81}\)[/tex] and then take the reciprocal due to the negative exponent:
[tex]\[ \left(\frac{16}{81}\right)^{-\frac{1}{4}} = \frac{1}{\left(\frac{16}{81}\right)^{\frac{1}{4}}} = \frac{1}{\frac{2}{3}} = \frac{3}{2} \][/tex]
So, [tex]\(\left(\frac{16}{81}\right)^{-\frac{1}{4}} = 1.5\)[/tex].
### n) [tex]\(\left(\frac{625}{81}\right)^{-0.5}\)[/tex]
First, take the square root of [tex]\(\frac{625}{81}\)[/tex] and then take the reciprocal due to the negative exponent:
[tex]\[ \left(\frac{625}{81}\right)^{-0.5} = \frac{1}{\left(\frac{625}{81}\right)^{0.5}} = \frac{1}{\frac{25}{9}} = \frac{9}{25} = 0.36 \][/tex]
So, [tex]\(\left(\frac{625}{81}\right)^{-0.5} = 0.36\)[/tex].
### j) [tex]\(8^{-\frac{1}{3}}\)[/tex]
Express 8 as [tex]\(2^3\)[/tex]:
[tex]\[ 8 = 2^3 \][/tex]
Next, take the cube root and then take the reciprocal due to the negative exponent:
[tex]\[ \left(2^3\right)^{-\frac{1}{3}} = 2^{3 \cdot -\frac{1}{3}} = 2^{-1} = \frac{1}{2} \][/tex]
So, [tex]\(8^{-\frac{1}{3}} = 0.5\)[/tex].
### o) [tex]\(\left(\frac{243}{32}\right)^{-0.4}\)[/tex]
Take the expression to the power of [tex]\(-0.4\)[/tex]:
[tex]\[ \left(\frac{243}{32}\right)^{-0.4} = \left(\frac{32}{243}\right)^{0.4} \][/tex]
Although this result was directly computed, its numerical value is approximately:
[tex]\[ 0.444444 \][/tex]
So, [tex]\(\left(\frac{243}{32}\right)^{-0.4} = 0.444444\)[/tex].
