Given that the derivative of a function is log, how to find the original function with a as the base and X as the true number? Given that the derivative of a function is log, how to find the original function with a as the base and X as the true number?

Given that the derivative of a function is log, how to find the original function with a as the base and X as the true number? Given that the derivative of a function is log, how to find the original function with a as the base and X as the true number?

First simplify the following with natural logarithm:
log_ a(x)=lnx/lna
‡ indefinite integral ∫ log_ a(x) dx
=(1/lna)∫lnx dx
=(1/lna)(xlnx-∫x dlnx)
=(1/lna)(xlnx-∫dx)
=(1/lna)(xlnx-x)+C
=xlnx/lna-xlna+C
=x[log_a(x)-lna]+C

(1) If the function y = log (a ^ 2-1 is the base) x is a true number and is a subtractive function on (0, positive infinity), then a belongs to? (2) What is the solution of inequality log (2) x < = - 1 / 2? Please write down the details of the process

X is a true number and is a subtractive function at (0, positive infinity), so 0

(text) if the function f (x) = log2 (2-ax) is a subtractive function on [0,1], then the value range of real number a is ___

∵ the function y = log2 (2-ax) monotonically decreases on [0, 1],
It is obtained that u = 2-ax is a subtractive function of X and is always positive on [0,1],
‡ a ＞ 0 and 2-A × 1 > 0, the solution is 0 < a < 2,
Therefore, the value range of a is 0 ＜ a ＜ 2
So the answer is: 0 < a < 2

If the function f (x) = log a is a function whose base (2-ax) is a true number and is a subtractive function of X in the interval [0,1], then the value range of A

Answer:
F (x) = loga (2-ax) is a subtractive function on interval [0,1]
1) 00, A1:
2-ax > 0 is a subtractive function on [0,1]
So: - A0
And: when x = 1, 2-ax = 2-A > 0, a

Log 2 (base) x ^ 2-ax + 3A (true number) is a subtractive function in (2, + ∞). Find the value of A

The base 2 is greater than 1, so log2x is an increasing function, so the monotonicity of log 2 (base) x ^ 2-ax + 3a is the same as the real number, so x ^ 2-ax + 3a is a decreasing function in (2, + ∞) and x ^ 2-ax + 3a is open upward, then it is an increasing function on the right side of the axis of symmetry, so it is impossible to be a decreasing function in (2, + ∞). If your topic is in (2, + ∞

How to find the derivative of Ln (x + 2)

Derivation of compound function
Let y = ln (x + 2), t = x + 2
Then y = LNT
y'=(lnt)'=1/t × t'=1/t × (x+2)'=1/(x+2) × 1=1/(x+2)