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The cubic power of y = x + log the logarithm of X with base 2 as the limit As in the title, Wrong, it's the whole derivative-
Log, the derivative of the logarithm of X base 2 is 1 / [x * ln (2)]
The derivative of x ^ 3 is 3x ^ 2
The entire derivative is 3x ^ 2 + 1 / [x * ln (2)]
If f (x) = 3 to the power of X - log with 2 as the base, the zero point of the logarithm of X is in the interval of A (- 2 / 5, - 2) B (- 2, - 1) C (1,2) d (2,2 / 5) solution
Answer: B
Resolution: base - x, that is, excluding answers C, D, logarithm definition field must be greater than zero
Replace x = - 2 into f (- 2) = 1 / 9-10
Then there must be zero in (- 1, - 2)
Multiple choice questions generally exclude the options qualitatively first, and the descendants are included in the calculation (Quantitative exclusion or selected options). The time spent in the examination will be reduced. Avoid excluding the options one by one
The 1 + log of 2 is based on 2, and the logarithm of 5 is Find out the specific calculation process
2^[1+log2(5)]
=2^[log2(2)+log2(5)]
=2^[log2(10)
=10
Log base 6 logarithm of 4 + log base 6 logarithm of 9-8 to the second power
Original formula = log6 (4 × 9) - (2 3) to the power of 2 / 3
=log6(6²)-2²
=2-4
=-2
Let a = log take 1 / 3 as the base logarithm of 2, B = log take 1 / 2 as the logarithm of 3, and C = 1 / 2 to the 0.3 power, and compare the sizes of a, B and C
The logarithm of log with 1 / 3 base is greater than that of log with 1 / 3 as base 3
The logarithm of log with 1 / 3 as base is greater than that with log with 1 / 2 as base 3
So 0 > a > B
And C > 0
So c > a > B
Solution equation: log2 (9 ^ X - 5) = log2 (3 ^ X - 2) + 2 Solve the equation log2 (9 ^ X - 5) = log2 (3 ^ X - 2) + 2 Hope there is a detailed process Note: 2 in log2 is the subscript of log
Log2 (9 ^ X - 5) = log2 (3 ^ X - 2) + 2 = log2 (4 * 3 ^ X-8) 9 ^ X-5 = 4 * 3 ^ X-8 (3 ^ x) ^ 2-5 = 4 * 3 ^ X-8 (3 ^ x) ^ 2-4 * 3 ^ x + 3 = 0 (3 ^ x-3) (3 ^ x-1) = 03 ^ x-3 = 0 or 3 ^ X-1 = x = 1 or x = 0, when x = 0, 9 ^ 0-5 = - 4, meaningless, x = 1
X ^ log2 (x + 2) = 8 to solve the logarithmic equation,
Take log2 on both sides at the same time
log2 x^log2(x+2)=log2 8
log2 (x+2) * log2 x=log2 8
log2 (x+2)x=log2 8
x(x+2)=8
X=2
Solve the equation log 2 (2 ^ x + 1) log2 [2 ^ (x + 1) + 2] = 2 find x the answer is 0 why thank you
Log 2 (2 ^ x + 1) log2 (2 ^ x + 1) log2 (2 ^ x + 1) log2 (2 ^ x + 1) log2 (2 ^ x + 1) log2 (2 ^ x + 1) log2 (2 ^ x + 1) log2 (2 ^ x + 1) log 2 (2 ^ x + 1) log 2 (2 ^ x + 1) = 2, set log2 (2 ^ x + 1) = 2, set log2 (2 ^ x + 1) = 2, set log2 (2 ^ x + 1) = x, X is simplified to x ^ 2 + X-2 = 0 x = - 2 or x = 1 because 2 ^ x + 1 > 1, so log 2 (2 ^ x + 1) > 2 (2 ^ x + 1) > 1 (2 ^ x + 1) > 2 (2 ^ x + 1) > 1, because of the 0, so you can't
Solve the equation log2 ^ (4 ^ x + 4) = x + log2 ^ [2 ^ (x + 1) - 3]
(4 ^ x + 4) = x + log2 ^ 2 ^ (x + 1) - 3] log2 ^ (4 ^ x + 4) - log2 ^ [2 ^ (x + 1) - 3] = xlog2 [(4 ^ x + 4) / (2 ^ (x + 1) - 3)] = x (4 ^ x + 4) / (2 ^ (x + 1) - 3) = 2 ^ X4 ^ x + 4 = 2 ^ x [2 ^ x + 1) - 3] 4 ^ x + 4 = 2 * 4 ^ x-3 * 2 ^ x-4 ^ x + 3 * 2 ^ x + 3 * 2 ^ x + 4 = 0, set 2 ^ x = x = x = x = x = 2 ^ x + 3 * 2 ^ x + 3 * 2 ^ x + 4 = 0, set 2 ^ x = 0, set 2 ^ x = 0, x = when t t t > 0, then - T 2 + 3 T + 4 = 0 t
Solving the equation log2 (4 ^ x + 1) = x + log2 (2 ^ (x + 3) - 6)
Log 2 (4 ^ x + 1) = x + log2 [2 ^ x + 3) - 6] log 2 (4 ^ x + 1) - log2 [2 ^ x + 3) - 6] = xlog2 [(4 ^ x + 1) / (2 ^ x × 8-6)] = x is 2 ^ x = (4 ^ x + 1) / (2 ^ x × 8-6) 2 ^ x (2 ^ x × 8-6) = 4 ^ x + 18 × 4 ^ X-6 × 2 ^ 2 ^ x = 4 ^ x + 18 × 4 ^ X-6 × 2 ^ X-1 = 4 ^ x + 17 × 4 ^ X-6 × 2 ^ X-1 = 1 = 0, make 2 ^ x = t, t, t, t, let 2 ^ x = t, let 2 ^ x = t, let 2 ^ x = t, let 2 ^ x = t, let then t ＞ 07t? - 6T -
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