Материалы по запросу: 2log2 3 1
Log2 (2x+1)=2log2 3-log2 (x-4)
Log2 (2x+1)=2log2 3-log2 (x-4)
Ответ на вопрос
First, use the properties of logarithms to simplify the equation:Log2 (2x+1) = 2log2 3 - log2 (x-4)
Log2 (2x+1) = log2 (3^2) - log2 (x-4)^1
Log2 (2x+1) = log2 (9) - log2 (x-4)Now, use the properties of logarithms to combine the logs on the right side:Log2 (2x+1) = log2 (9/(x-4))Since the bases of the logarithms are the same, we can set the arguments equal to each other:2x + 1 = 9/(x-4)Now, solve for x:2x + 1 = 9/(x-4)
2x(x-4) + (x-4) = 9
2x^2 - 8x + x - 4 = 9
2x^2 - 7x - 13 = 0Now, you can use the quadratic formula to solve for x:x = (7 ± sqrt(7^2 - 42(-13))) / (2*2)
x = (7 ± sqrt(49 + 104)) / 4
x = (7 ± sqrt(153)) / 4So, the solutions for x are:x = (7 + sqrt(153)) / 4
x = (7 - sqrt(153)) / 4
Еще
72
1
32^2/3 =3^-4 = log32 43 - 2log2 1 =
32^2/3 =3^-4 = log32 43 - 2log2 1 =
Ответ на вопрос
1/3Explanation:
32^(2/3) = (2^5)^(2/3) = 2^(10/3) = 2^3 = 8
3^(-4) = 1/3^4 = 1/81
log32 43 = log2 43 / log2 32 = log2 43 / 5
2log2 1 = 2(0) = 0Therefore, the final result is:
(8) / (1/81) - log2 43 / 5 - 0 = 648 - (log2 43) / 5
Еще
157
1
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