LG2 = 0.3010, LG1 / 2

LG2 = 0.3010, LG1 / 2

The original formula is: lg1-lg2 = - 0.3010 LG1 = 0.lg2 = 0.3010

How to find the derivative of [f (x) = x power of a - x * LNA]

F'(X)=a^x㏑a-㏑a
Mingjiao answers for you,
If it can help you, I hope you can click (satisfactory answer)
If not, please ask, I will try my best to help you solve it~
The answer is not easy. If you are not satisfied, please understand~

Given f (x) = 2x + 2-x, if f (a) = 3, then f (2a)=______ ．

∵ f (a) = 3 = 2A + 2-A, ∵ f (2a) = 22a + 2-2a = (2a + 2-A) 2-2 = 32-2 = 7

3x (1 - (x + 1)) of (x to the power of 2-1)

=3x/(x+1)(x-1)÷(x+1-1)/(x+1)
=3x/(x+1)(x-1)×(x+1)/x
=3/(x-1)

Let f (x) = 3x + 3x-8, in the process of finding the approximate solution of equation 3x + 3x-8 = 0 in X ∈ (1,3) by dichotomy, take the midpoint of interval x0 = 2, then the next rooted interval is ()
A. (1,2) B. (2,3) C. (1,2) or (2,3) d. cannot be determined

∵ f (1) = 31 + 3 × 1-8 = - 2 ＜ 0, f (3) = 33 + 3 × 3-8 = 28 ＞ 0, f (2) = 32 + 3 × 2-8 = 7 ＞ 0, the next rooted interval of ∵ f (1) f (2) ＜ 0 and ∵ f (x) = 0 is (1,2)

It is known that the equation 3x with respect to X to the power of N + 2 + 8 = 0 is an equation of one variable to the power of 1, and the value of N can be obtained

Because it's a linear equation of one variable
So n + 2 = 1
n=-1

If f (x) = LNX, then f (2) =?
I'm sorry. I got the wrong number to find the value of F (2). Excuse me?

Let x ^ n = t (x ^ n is the nth power of x)
x=t^(1/n)
f(t)=ln[t^(1/n)]
=(lnt)/n
That is f (x) = (LNX) / n
f(2)=(ln2)/n
That's it

If f '(1) = - 4, then n =?
Is the original function over (1, - 4)?
If it is regarded as a derivative function, it will be over (1,0)?
Isn't that the same as the extreme value of F '(x) at 1 is - 4?

You should remember that the extreme value of function f (x) means that when x = x1, its derivative f '(x) = 0, that is, f' (x1) = 0, then f (x1) is the maximum or minimum value of function f (x)
By analogy, the value of F '(x) level you said is. When x takes x 1, the derivative function of F' (x) f '' (x 1) = 0. Then f '(x 1) is the maximum or minimum of F' (x). Obviously, it has nothing to do with this problem
And there's no such thing as f '(x) at 1
F '(1) = - 4 means that f' (x) passes (1, - 4) instead of the original function passing (1, - 4)
In this problem, we can get n = - 4 by taking the point (1, - 4) into f '(x)
More gnawing gnawing concept, the concept can be clear to do the problem ah
I hope my explanation is clear enough for you to understand

Are f (x) = 1 and G (x) = x equal to the power 0

In G (x) = x0, X cannot take 0

F (x) = 1 and G (x) = the zeroth power of X

The two groups of functions are not equal, and the functions are equal conditionally: first, the domain of definition is the same; second, the expression is the same; the domain of definition of the two functions above is different; the domain of definition of the first function is R; and the second is that x is not equal to 0