find the value of x
![{2}^{5x + 2} \div {2}^{x} = \sqrt[5]{2 ^{20} }](https://tex.z-dn.net/?f=+%7B2%7D%5E%7B5x+%2B+2%7D++%5Cdiv++%7B2%7D%5E%7Bx%7D++%3D++%5Csqrt%5B5%5D%7B2+%5E%7B20%7D+%7D+ "{2}^{5x + 2} \div {2}^{x} = \sqrt[5]{2 ^{20} }")
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2^(5x + 2) ÷ 2^x = sqrt[5]{20^20}
2^(5x + 2) × 1/2^x = sqrt[5]{20^20}
2^(5x + 2)/2^x = 20^{20/5}
[ we know, x^m/x^n = x^(m-n) use it here, ]
2^(5x + 2 - x) = 20^4
2^(4x + 2) = 2^4
[ we know, x^m = x^n then, m = n ]
(4x + 2) = 4
4x = 4 - 2 = 2
x = 2/4 = 1/2
2^(5x + 2) × 1/2^x = sqrt[5]{20^20}
2^(5x + 2)/2^x = 20^{20/5}
[ we know, x^m/x^n = x^(m-n) use it here, ]
2^(5x + 2 - x) = 20^4
2^(4x + 2) = 2^4
[ we know, x^m = x^n then, m = n ]
(4x + 2) = 4
4x = 4 - 2 = 2
x = 2/4 = 1/2
thindsaab:
sorry but I don't understand
Your question is 2^(5x +2) ÷ 2^x = sqrt[5]{2^20} .we know, ⁿ√x = x^(1/n) so, sqrt[5]{2^20} = (2^20)^(1/5) = (2)^(20×1/5) = 2^4
Now, 2^(5x + 2) ÷ 2^x = 2^4 we know, 1÷2 means 1/2 . Use this here , 2^(5x+2)÷2^x = 2^4 , 2^(5x+2)/2^x=2^4 ,we know, according to rule , x^m/x^n =x^(m-n) use it here, 2^(5x+2-x)=2^4
And now, finally we know, x^m = x^n then, m = n , so, 2^(4x +2) = 2^4 then, (4x +2) = 4 , x = 1/2
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