What is 4piar^2*((1/2)^{1/4}(z/r)^{1/2}*T)*T^4 ?

The solution to 4piar^2*((1/2)^{1/4}(z/r)^{1/2}*T)*T^4 is 4pi(1/2)^{1/4}ar^2(z/r)^{1/2}T^5

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simplify 4piar^2((1/2)^{1/4}(z/r)^{1/2}T)*T^4

simplify

4 πa r^{2} \cdot ((\frac{1}{2})^{\frac{1}{4}} (\frac{z}{r})^{\frac{1}{2}} \cdot T) \cdot T^{4}

4 π (\frac{1}{2})^{\frac{1}{4}} a r^{2} (\frac{z}{r})^{\frac{1}{2}} T^{5}

Solution steps

4 πa r^{2} ((\frac{1}{2})^{\frac{1}{4}} (\frac{z}{r})^{\frac{1}{2}} T) T^{4}

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

T T^{4} = T^{1 + 4}

= 4 πa r^{2} (\frac{1}{2})^{\frac{1}{4}} (\frac{z}{r})^{\frac{1}{2}} T^{1 + 4}

Add the numbers:

1 + 4 = 5

= 4 πa r^{2} (\frac{1}{2})^{\frac{1}{4}} (\frac{z}{r})^{\frac{1}{2}} T^{5}

(3 a^{2} b^{7}) (7 a^{3} b^{25})

(5 z)^{2}

a^{2} (b + c)^{2}

- 3 a^{2} b^{2} \cdot (b^{5} - 2 a b^{2})

\frac{5}{5} \cdot \frac{5}{5}

What is 4piar^2*((1/2)^{1/4}(z/r)^{1/2}*T)*T^4 ?
The solution to 4piar^2*((1/2)^{1/4}(z/r)^{1/2}*T)*T^4 is 4pi(1/2)^{1/4}ar^2(z/r)^{1/2}T^5