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Первое действие логарифмируем по основанию 8 левую и правую часть. И далее получить ответ гораздо проще. В несколько действий. 2/3*(логарифм по основанию 8 от 5 -1)
What about the complex roots?8^(x + 1) = 508^x * 8^1 = 508^x * 8 / 8 = 50 / 88^x * 1 = 2 * 5^2 / 2^38^x = 5^2 / 2^2(2^3)^x = (5 / 2)^22^(3 * x) = ([(5 / 2)^2]^[1 / 3])^32^(x * 3) = ([5 / 2]^[2 * (1 / 3)])^3(2^x)^3 = ([5 / 2]^[2 / 3])^3Let a = 2^x, and b = (5 / 2)^(2 / 3)(2^x)^3 = ([5 / 2]^[2 / 3])^3=> a^3 = b^3=> a^3 - b^3 = b^3 - b^3=> a^3 - b^3 = 0=> (a - b)(a^2 + ab + b^2) = 0=> (a - b)(1 * a^2 + b * a + b^2) = 0=> a - b = 0, or 1 * a^2 + b * a + b^2 = 0Suppose a - b = 0a - b = 0a - b + b = 0 + ba = bRemember, a = 2^x, and b = (5 / 2)^(2 / 3)2^x = (5 / 2)^(2 / 3)ln(2^x) = ln([5 / 2]^[2 / 3])x * ln(2) = ln([5 / 2]^[2 / 3])x * ln(2) / ln(2) = ln([5 / 2]^[2 / 3]) / ln(2)x * log_2(2) = log_2([5 / 2]^[2 / 3])x * 1 = (2 / 3) * log_2(5 / 2)x = (2 / 3) * (log_2[5] - log_2[2])x = (2 / 3) * (log_2[5] - 1)x = 2 * (log_2[5] - 1) / 3x = 2 * log_2(5) / 3 - 2 * 1 / 3x = 2 * log_2(5) / 3 - 2 / 3Suppose 1 * a^2 + b * a + b^2 = 01 * a^2 + b * a + b^2 = 0a = (-b +/- sqrt[b^2 - 4 * 1 * b^2]) / (2 * 1)a = (-b +/- sqrt[1 * b^2 - 4 * b^2]) / (2)a = (-b +/- sqrt[(1 - 4) * b^2]) / 2a = (-b +/- sqrt[(-3) * b^2]) / 2a = (-b +/- sqrt[(-1) * 3 * b^2]) / 2a = (-b +/- sqrt[-1] * sqrt[3] * sqrt[b^2]) / 2a = (-b +/- i * sqrt[3] * b) / 2a = b * (-1 +/- i * sqrt[3] * 1) / 2Remember, a = 2^x, and b = (5 / 2)^(2 / 3)2^x = (5 / 2)^(2 / 3) * (-1 +/- i * 3^[1 / 2]) / 2^12^x = 2^(-1) * (5 / 2)^(2 / 3) * (-1 +/- i * 3^[1 / 2])2^x = 2^(-1) * 5^(2 / 3) * (2^[-1])^(2 / 3) * (-1 +/- i * 3^[1 / 2])2^x = 2^(-1) * (2^[-1])^(2 / 3) * 5^(2 / 3) * (-1 +/- i * 3^[1 / 2])2^x = 2^(-1) * 2^([-1] * [2 / 3]) * 5^(2 / 3) * (-1 +/- i * 3^[1 / 2])2^x = 2^(-1) * 2^(-2 / 3) * 5^(2 / 3) * (-1 +/- i * 3^[1 / 2])2^x = 2^(-1 - [2 / 3]) * 5^(2 / 3) * (-1 +/- i * 3^[1 / 2])2^x = 2^(-5 / 3) * 5^(2 / 3) * (-1 +/- i * 3^[1 / 2])ln(2^x) = ln(2^[-5 / 3] * 5^[2 / 3] * [-1 +/- i * 3^(1 / 2)])x * ln(2) = ln(2^[-5 / 3] * 5^[2 / 3] * [-1 +/- i * 3^(1 / 2)])x * ln(2) = ln(2^[-5 / 3] * 5^[2 / 3] * [-1 + i * 3^(1 / 2)]), or x * ln(2) = ln(2^[-5 / 3] * 5^[2 / 3] * [-1 - i * 3^(1 / 2)])Suppose x * ln(2) = ln(2^[-5 / 3] * 5^[2 / 3] * [-1 + i * 3^(1 / 2)])x * ln(2) = ln(2^[-5 / 3] * 5^[2 / 3] * [i^2 + i * 3^(1 / 2)])x * ln(2) = ln(2^[-5 / 3] * 5^[2 / 3] * i * [i + 1 * 3^(1 / 2)])x * ln(2) / ln(2) = ln(2^[-5 / 3] * 5^[2 / 3] * i * [i + 3^(1 / 2)]) / ln(2) x * log_2(2) = ln(2^[-5 / 3]) / ln(2) + ln(5^[2 / 3]) / ln(2) + ln(i) / ln(2) + ln(i + 3^[1 / 2]) / ln(2) x * log_2(2) = ln(2^[-5 / 3]) / ln(2) + ln(5^[2 / 3]) / ln(2) + ln(e^[i * tau / 4]) / ln(2) + ln(i + 3^[1 / 2]) / ln(2)x * 1 = (-5 / 3) * ln(2) / ln(2) + (2 / 3) * ln(5) / ln(2) + (i * tau / 4) * ln(e) / ln(2) + ln(i + 3^[1 / 2]) / ln(2)x = (-5 / 3) * log_2(2) + (2 / 3) * log_2(5) + (i * tau / 4) * 1 / ln(2) + log_2(i + 3^[1 / 2])x = (-5 / 3) * 1 + (2 / 3) * log_2(5) + i * tau / (4 * ln[2]) + log_2(i + 3^[1 / 2])x = (-5 / 3) + 2 * log_2(5) / 3 + i * tau / (4 * ln[2]) + log_2(i + 3^[1 / 2])Suppose x * ln(2) = ln(2^[-5 / 3] * 5^[2 / 3] * [-1 - i * 3^(1 / 2)])x * ln(2) = ln(2^[-5 / 3] * 5^[2 / 3] * [-1 - 1 * i * 3^(1 / 2)])x * ln(2) = ln(2^[-5 / 3] * 5^[2 / 3] * [-1] * [1 + i * 3^(1 / 2)])x * ln(2) = ln(2^[-5 / 3] * 5^[2 / 3] * i^2 * [1 + i * 3^(1 / 2)])x * ln(2) / ln(2) = ln(2^[-5 / 3] * 5^[2 / 3] * i^2 * [1 + i * 3^(1 / 2)]) / ln(2)x * log_2(2) = ln(2^[-5 / 3]) / ln(2) + ln(5^[2 / 3]) / ln(2) + ln(i^2) / ln(2) + ln(1 + i * 3^[1 / 2]) / ln(2)x * 1 = ln(2^[-5 / 3]) / ln(2) + ln(5^[2 / 3]) / ln(2) + ln(e^[i * tau / 2]) / ln(2) + ln(1 + i * 3^[1 / 2]) / ln(2)x = (-5 / 3) * ln(2) / ln(2) + (2 / 3) * ln(5) / ln(2) + (i * tau / 2) * ln(e) / ln(2) + ln(1 + i * 3^[1 / 2]) / ln(2)x = (-5 / 3) * log_2(2) + (2 / 3) * log_2(5) + (i * tau / 2) * 1 / ln(2) + log_2(1 + i * 3^[1 / 2])x = (-5 / 3) * 1 + 2 * log_2(5) / 3 + i * tau / (2 * ln[2]) + log_2(1 + i * 3^[1 / 2])x = (-5 / 3) + 2 * log_2(5) / 3 + i * tau / (2 * ln[2]) + log_2(1 + i * 3^[1 / 2])x1 = -(2 / 3) + 2 * log_2(5) / 3x2 = -(5 / 3) + 2 * log_2(5) / 3 + i * tau / (4 * ln[2]) + log_2(i + 3^[1 / 2])x3 = -(5 / 3) + 2 * log_2(5) / 3 + i * tau / (2 * ln[2]) + log_2(1 + i * 3^[1 / 2])
I was super confused until I realized the "+1" was also in the exponent.
8^x.8=50 Or 8^x.8=50Or 8^x=50/8Or xlog8=log(50/8) Or x={( log50- log8) /log8}(এর পর হয়তো আরো ভাঙা যাবে)
log8_6.25
8^^(x+1)=50(x+1)*log(8)=log(50)(x+1)=log(50)/log(8)x+1=1,699/0,903=1,882x=1,882-1=0.882x=0.8828^^(0,882+1)=508^^1.882=50.07
You can just take the logarithm of base eight and then subject one. It shouldn't be this hard
8^x×8=508^x=50/8=25/4x=(log25-log4)/log8=⅔((log5/log2)-1)x≈⅔((0.699/0.301)-1)≈796/903≈0.8815Deviation is about 0.00028^(0.8815+1)≈50.02
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Первое действие логарифмируем по основанию 8 левую и правую часть. И далее получить ответ гораздо проще. В несколько действий. 2/3*(логарифм по основанию 8 от 5 -1)
What about the complex roots?
8^(x + 1) = 50
8^x * 8^1 = 50
8^x * 8 / 8 = 50 / 8
8^x * 1 = 2 * 5^2 / 2^3
8^x = 5^2 / 2^2
(2^3)^x = (5 / 2)^2
2^(3 * x) = ([(5 / 2)^2]^[1 / 3])^3
2^(x * 3) = ([5 / 2]^[2 * (1 / 3)])^3
(2^x)^3 = ([5 / 2]^[2 / 3])^3
Let a = 2^x, and b = (5 / 2)^(2 / 3)
(2^x)^3 = ([5 / 2]^[2 / 3])^3
=> a^3 = b^3
=> a^3 - b^3 = b^3 - b^3
=> a^3 - b^3 = 0
=> (a - b)(a^2 + ab + b^2) = 0
=> (a - b)(1 * a^2 + b * a + b^2) = 0
=> a - b = 0, or 1 * a^2 + b * a + b^2 = 0
Suppose a - b = 0
a - b = 0
a - b + b = 0 + b
a = b
Remember, a = 2^x, and b = (5 / 2)^(2 / 3)
2^x = (5 / 2)^(2 / 3)
ln(2^x) = ln([5 / 2]^[2 / 3])
x * ln(2) = ln([5 / 2]^[2 / 3])
x * ln(2) / ln(2) = ln([5 / 2]^[2 / 3]) / ln(2)
x * log_2(2) = log_2([5 / 2]^[2 / 3])
x * 1 = (2 / 3) * log_2(5 / 2)
x = (2 / 3) * (log_2[5] - log_2[2])
x = (2 / 3) * (log_2[5] - 1)
x = 2 * (log_2[5] - 1) / 3
x = 2 * log_2(5) / 3 - 2 * 1 / 3
x = 2 * log_2(5) / 3 - 2 / 3
Suppose 1 * a^2 + b * a + b^2 = 0
1 * a^2 + b * a + b^2 = 0
a = (-b +/- sqrt[b^2 - 4 * 1 * b^2]) / (2 * 1)
a = (-b +/- sqrt[1 * b^2 - 4 * b^2]) / (2)
a = (-b +/- sqrt[(1 - 4) * b^2]) / 2
a = (-b +/- sqrt[(-3) * b^2]) / 2
a = (-b +/- sqrt[(-1) * 3 * b^2]) / 2
a = (-b +/- sqrt[-1] * sqrt[3] * sqrt[b^2]) / 2
a = (-b +/- i * sqrt[3] * b) / 2
a = b * (-1 +/- i * sqrt[3] * 1) / 2
Remember, a = 2^x, and b = (5 / 2)^(2 / 3)
2^x = (5 / 2)^(2 / 3) * (-1 +/- i * 3^[1 / 2]) / 2^1
2^x = 2^(-1) * (5 / 2)^(2 / 3) * (-1 +/- i * 3^[1 / 2])
2^x = 2^(-1) * 5^(2 / 3) * (2^[-1])^(2 / 3) * (-1 +/- i * 3^[1 / 2])
2^x = 2^(-1) * (2^[-1])^(2 / 3) * 5^(2 / 3) * (-1 +/- i * 3^[1 / 2])
2^x = 2^(-1) * 2^([-1] * [2 / 3]) * 5^(2 / 3) * (-1 +/- i * 3^[1 / 2])
2^x = 2^(-1) * 2^(-2 / 3) * 5^(2 / 3) * (-1 +/- i * 3^[1 / 2])
2^x = 2^(-1 - [2 / 3]) * 5^(2 / 3) * (-1 +/- i * 3^[1 / 2])
2^x = 2^(-5 / 3) * 5^(2 / 3) * (-1 +/- i * 3^[1 / 2])
ln(2^x) = ln(2^[-5 / 3] * 5^[2 / 3] * [-1 +/- i * 3^(1 / 2)])
x * ln(2) = ln(2^[-5 / 3] * 5^[2 / 3] * [-1 +/- i * 3^(1 / 2)])
x * ln(2) = ln(2^[-5 / 3] * 5^[2 / 3] * [-1 + i * 3^(1 / 2)]), or x * ln(2) = ln(2^[-5 / 3] * 5^[2 / 3] * [-1 - i * 3^(1 / 2)])
Suppose x * ln(2) = ln(2^[-5 / 3] * 5^[2 / 3] * [-1 + i * 3^(1 / 2)])
x * ln(2) = ln(2^[-5 / 3] * 5^[2 / 3] * [i^2 + i * 3^(1 / 2)])
x * ln(2) = ln(2^[-5 / 3] * 5^[2 / 3] * i * [i + 1 * 3^(1 / 2)])
x * ln(2) / ln(2) = ln(2^[-5 / 3] * 5^[2 / 3] * i * [i + 3^(1 / 2)]) / ln(2)
x * log_2(2) = ln(2^[-5 / 3]) / ln(2) + ln(5^[2 / 3]) / ln(2) + ln(i) / ln(2) + ln(i + 3^[1 / 2]) / ln(2)
x * log_2(2) = ln(2^[-5 / 3]) / ln(2) + ln(5^[2 / 3]) / ln(2) + ln(e^[i * tau / 4]) / ln(2) + ln(i + 3^[1 / 2]) / ln(2)
x * 1 = (-5 / 3) * ln(2) / ln(2) + (2 / 3) * ln(5) / ln(2) + (i * tau / 4) * ln(e) / ln(2) + ln(i + 3^[1 / 2]) / ln(2)
x = (-5 / 3) * log_2(2) + (2 / 3) * log_2(5) + (i * tau / 4) * 1 / ln(2) + log_2(i + 3^[1 / 2])
x = (-5 / 3) * 1 + (2 / 3) * log_2(5) + i * tau / (4 * ln[2]) + log_2(i + 3^[1 / 2])
x = (-5 / 3) + 2 * log_2(5) / 3 + i * tau / (4 * ln[2]) + log_2(i + 3^[1 / 2])
Suppose x * ln(2) = ln(2^[-5 / 3] * 5^[2 / 3] * [-1 - i * 3^(1 / 2)])
x * ln(2) = ln(2^[-5 / 3] * 5^[2 / 3] * [-1 - 1 * i * 3^(1 / 2)])
x * ln(2) = ln(2^[-5 / 3] * 5^[2 / 3] * [-1] * [1 + i * 3^(1 / 2)])
x * ln(2) = ln(2^[-5 / 3] * 5^[2 / 3] * i^2 * [1 + i * 3^(1 / 2)])
x * ln(2) / ln(2) = ln(2^[-5 / 3] * 5^[2 / 3] * i^2 * [1 + i * 3^(1 / 2)]) / ln(2)
x * log_2(2) = ln(2^[-5 / 3]) / ln(2) + ln(5^[2 / 3]) / ln(2) + ln(i^2) / ln(2) + ln(1 + i * 3^[1 / 2]) / ln(2)
x * 1 = ln(2^[-5 / 3]) / ln(2) + ln(5^[2 / 3]) / ln(2) + ln(e^[i * tau / 2]) / ln(2) + ln(1 + i * 3^[1 / 2]) / ln(2)
x = (-5 / 3) * ln(2) / ln(2) + (2 / 3) * ln(5) / ln(2) + (i * tau / 2) * ln(e) / ln(2) + ln(1 + i * 3^[1 / 2]) / ln(2)
x = (-5 / 3) * log_2(2) + (2 / 3) * log_2(5) + (i * tau / 2) * 1 / ln(2) + log_2(1 + i * 3^[1 / 2])
x = (-5 / 3) * 1 + 2 * log_2(5) / 3 + i * tau / (2 * ln[2]) + log_2(1 + i * 3^[1 / 2])
x = (-5 / 3) + 2 * log_2(5) / 3 + i * tau / (2 * ln[2]) + log_2(1 + i * 3^[1 / 2])
x1 = -(2 / 3) + 2 * log_2(5) / 3
x2 = -(5 / 3) + 2 * log_2(5) / 3 + i * tau / (4 * ln[2]) + log_2(i + 3^[1 / 2])
x3 = -(5 / 3) + 2 * log_2(5) / 3 + i * tau / (2 * ln[2]) + log_2(1 + i * 3^[1 / 2])
I was super confused until I realized the "+1" was also in the exponent.
8^x.8=50
Or
8^x.8=50
Or 8^x=50/8
Or xlog8=log(50/8)
Or x={( log50- log8) /log8}
(এর পর হয়তো আরো ভাঙা যাবে)
log8_6.25
8^^(x+1)=50
(x+1)*log(8)=log(50)
(x+1)=log(50)/log(8)
x+1=1,699/0,903=1,882
x=1,882-1=0.882
x=0.882
8^^(0,882+1)=50
8^^1.882=50.07
You can just take the logarithm of base eight and then subject one. It shouldn't be this hard
8^x×8=50
8^x=50/8=25/4
x=(log25-log4)/log8=⅔((log5/log2)-1)
x≈⅔((0.699/0.301)-1)≈796/903≈0.8815
Deviation is about 0.0002
8^(0.8815+1)≈50.02