Answer :
Sure, let's solve the given expressions step-by-step for each part.
a) [tex]\(\left(2^2\right)^3\)[/tex]
First, calculate [tex]\(2^2\)[/tex]:
[tex]\[ 2^2 = 4 \][/tex]
Then, raise this result to the power 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, calculate [tex]\(3^6\)[/tex]:
[tex]\[ 3^6 = 729 \][/tex]
Then, take 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, calculate [tex]\(5^2\)[/tex]:
[tex]\[ 5^2 = 25 \][/tex]
Then, take the square root:
[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, calculate [tex]\(7^{-3}\)[/tex]:
[tex]\[ 7^{-3} = \frac{1}{7^3} = \frac{1}{343} \][/tex]
Then, take the cube root:
[tex]\[ \sqrt[3]{\frac{1}{343}} = \frac{1}{\sqrt[3]{343}} = \frac{1}{7} \][/tex]
So, [tex]\(\left(7^{-3}\right)^{\frac{1}{3}} \approx 0.14285714285714288\)[/tex].
f) [tex]\(4^{\frac{1}{2}}\)[/tex]
Take 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]
Take 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]
First, express 8 as a power of 2:
[tex]\[ 8 = 2^3 \][/tex]
Then, [tex]\(8^{\frac{2}{3}}\)[/tex] becomes:
[tex]\[ (2^3)^{\frac{2}{3}} = 2^{3 \cdot \frac{2}{3}} = 2^2 = 4 \][/tex]
So, [tex]\(8^{\frac{2}{3}} \approx 3.9999999999999996\)[/tex].
i) [tex]\(16^{\frac{1}{4}}\)[/tex]
First, calculate [tex]\(16\)[/tex] as a power of [tex]\(2\)[/tex]:
[tex]\[ 16 = 2^4 \][/tex]
Then, take the 4th root:
[tex]\[ \sqrt[4]{16} = \sqrt[4]{2^4} = 2 \][/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 fraction inside:
[tex]\[ \frac{2^2}{3^2} = \frac{4}{9} \][/tex]
Now, raise to the power [tex]\(-\frac{1}{2}\)[/tex]:
[tex]\[ \left(\frac{4}{9}\right)^{-\frac{1}{2}} = \left(\frac{9}{4}\right)^{\frac{1}{2}} = \sqrt{\frac{9}{4}} = \frac{3}{2} \][/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 a power:
[tex]\[ 81 = 3^4 \][/tex]
Then, calculate:
[tex]\[ 81^{-\frac{3}{4}} = (3^4)^{-\frac{3}{4}} = 3^{-3} = \frac{1}{3^3} = \frac{1}{27} \][/tex]
So, [tex]\(81^{-\frac{3}{4}} \approx 0.037037037037037035\)[/tex].
l) [tex]\(\left(\frac{16}{25}\right)^{\frac{1}{2}}\)[/tex]
Take the square root of the fraction:
[tex]\[ \sqrt{\frac{16}{25}} = \frac{\sqrt{16}}{\sqrt{25}} = \frac{4}{5} \][/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]
Take the reciprocal of the fraction and then the 4th root:
[tex]\[ \left(\frac{16}{81}\right)^{-\frac{1}{4}} = \left(\frac{81}{16}\right)^{\frac{1}{4}} = \frac{\sqrt[4]{81}}{\sqrt[4]{16}} = \frac{3}{2} \][/tex]
So, [tex]\(\left(\frac{16}{81}\right)^{-\frac{1}{4}} = 1.5\)[/tex].
n) [tex]\(\left(\frac{625}{81}\right)^{-0.5}\)[/tex]
Take the reciprocal of the fraction and then the square root:
[tex]\[ \left(\frac{625}{81}\right)^{-0.5} = \left(\frac{81}{625}\right)^{0.5} = \sqrt{\frac{81}{625}} = \frac{9}{25} \][/tex]
So, [tex]\(\left(\frac{625}{81}\right)^{-0.5} = 0.36\)[/tex].
j) [tex]\(8^{-\frac{1}{3}}\)[/tex]
Express 8 as a power:
[tex]\[ 8 = 2^3 \][/tex]
Then, calculate the cube root and apply the negative exponent:
[tex]\[ 8^{-\frac{1}{3}} = (2^3)^{-\frac{1}{3}} = 2^{-1} = \frac{1}{2} \][/tex]
So, [tex]\(8^{-\frac{1}{3}} = 0.5\)[/tex].
o) [tex]\(\left(\frac{243}{32}\right)^{-0.4}\)[/tex]
First, raise the fraction to the power [tex]\(-0.4\)[/tex]:
[tex]\[ \left(\frac{243}{32}\right)^{-0.4} \approx 0.4444444444444444 \][/tex]
So, [tex]\(\left(\frac{243}{32}\right)^{-0.4} \approx 0.4444444444444444\)[/tex].
a) [tex]\(\left(2^2\right)^3\)[/tex]
First, calculate [tex]\(2^2\)[/tex]:
[tex]\[ 2^2 = 4 \][/tex]
Then, raise this result to the power 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, calculate [tex]\(3^6\)[/tex]:
[tex]\[ 3^6 = 729 \][/tex]
Then, take 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, calculate [tex]\(5^2\)[/tex]:
[tex]\[ 5^2 = 25 \][/tex]
Then, take the square root:
[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, calculate [tex]\(7^{-3}\)[/tex]:
[tex]\[ 7^{-3} = \frac{1}{7^3} = \frac{1}{343} \][/tex]
Then, take the cube root:
[tex]\[ \sqrt[3]{\frac{1}{343}} = \frac{1}{\sqrt[3]{343}} = \frac{1}{7} \][/tex]
So, [tex]\(\left(7^{-3}\right)^{\frac{1}{3}} \approx 0.14285714285714288\)[/tex].
f) [tex]\(4^{\frac{1}{2}}\)[/tex]
Take 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]
Take 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]
First, express 8 as a power of 2:
[tex]\[ 8 = 2^3 \][/tex]
Then, [tex]\(8^{\frac{2}{3}}\)[/tex] becomes:
[tex]\[ (2^3)^{\frac{2}{3}} = 2^{3 \cdot \frac{2}{3}} = 2^2 = 4 \][/tex]
So, [tex]\(8^{\frac{2}{3}} \approx 3.9999999999999996\)[/tex].
i) [tex]\(16^{\frac{1}{4}}\)[/tex]
First, calculate [tex]\(16\)[/tex] as a power of [tex]\(2\)[/tex]:
[tex]\[ 16 = 2^4 \][/tex]
Then, take the 4th root:
[tex]\[ \sqrt[4]{16} = \sqrt[4]{2^4} = 2 \][/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 fraction inside:
[tex]\[ \frac{2^2}{3^2} = \frac{4}{9} \][/tex]
Now, raise to the power [tex]\(-\frac{1}{2}\)[/tex]:
[tex]\[ \left(\frac{4}{9}\right)^{-\frac{1}{2}} = \left(\frac{9}{4}\right)^{\frac{1}{2}} = \sqrt{\frac{9}{4}} = \frac{3}{2} \][/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 a power:
[tex]\[ 81 = 3^4 \][/tex]
Then, calculate:
[tex]\[ 81^{-\frac{3}{4}} = (3^4)^{-\frac{3}{4}} = 3^{-3} = \frac{1}{3^3} = \frac{1}{27} \][/tex]
So, [tex]\(81^{-\frac{3}{4}} \approx 0.037037037037037035\)[/tex].
l) [tex]\(\left(\frac{16}{25}\right)^{\frac{1}{2}}\)[/tex]
Take the square root of the fraction:
[tex]\[ \sqrt{\frac{16}{25}} = \frac{\sqrt{16}}{\sqrt{25}} = \frac{4}{5} \][/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]
Take the reciprocal of the fraction and then the 4th root:
[tex]\[ \left(\frac{16}{81}\right)^{-\frac{1}{4}} = \left(\frac{81}{16}\right)^{\frac{1}{4}} = \frac{\sqrt[4]{81}}{\sqrt[4]{16}} = \frac{3}{2} \][/tex]
So, [tex]\(\left(\frac{16}{81}\right)^{-\frac{1}{4}} = 1.5\)[/tex].
n) [tex]\(\left(\frac{625}{81}\right)^{-0.5}\)[/tex]
Take the reciprocal of the fraction and then the square root:
[tex]\[ \left(\frac{625}{81}\right)^{-0.5} = \left(\frac{81}{625}\right)^{0.5} = \sqrt{\frac{81}{625}} = \frac{9}{25} \][/tex]
So, [tex]\(\left(\frac{625}{81}\right)^{-0.5} = 0.36\)[/tex].
j) [tex]\(8^{-\frac{1}{3}}\)[/tex]
Express 8 as a power:
[tex]\[ 8 = 2^3 \][/tex]
Then, calculate the cube root and apply the negative exponent:
[tex]\[ 8^{-\frac{1}{3}} = (2^3)^{-\frac{1}{3}} = 2^{-1} = \frac{1}{2} \][/tex]
So, [tex]\(8^{-\frac{1}{3}} = 0.5\)[/tex].
o) [tex]\(\left(\frac{243}{32}\right)^{-0.4}\)[/tex]
First, raise the fraction to the power [tex]\(-0.4\)[/tex]:
[tex]\[ \left(\frac{243}{32}\right)^{-0.4} \approx 0.4444444444444444 \][/tex]
So, [tex]\(\left(\frac{243}{32}\right)^{-0.4} \approx 0.4444444444444444\)[/tex].
