I. Exponential concept and logarithmic concept:
The concept of exponent is extended from the concept of power. Multiply the same factor by a...a (n) =an to find the power, where n is a positive integer. Starting from junior high school, N is first extended to all integers; Then the power sum root is unified and extended to rational exponent; Finally, the concept of exponent holds in the range of real numbers.
Euler pointed out: "Logarithmic source is exponential". Generally speaking, if a (a >; 0, a≠ 1) whose b power is equal to n, that is, ab=N, then this number b is called the logarithm of n with a base, and it is recorded as Logan = B.
Where a is called the base of logarithm and n is called real number.
Ab=N and b=logaN are a pair of equivalent formulas, where A is a given normal number that is not equal to 1. When b is given to n, it is an exponential operation, and when n is given to b, it is a logarithmic operation. The reciprocal operation of exponential operation and logarithmic operation.
Second, the essence of exponential operation and logarithmic operation
1. Exponential operation has three main properties:
ax ay=ax+y，(ax)y=axy，(ab)x = ax bx(a & gt； 0，a≠ 1，b & gt0，b≠ 1)
2. There are three logarithmic algorithms (attributes):
( 1)loga(MN)=logaM+logaN
(2)logaM/N=logaM-logaN
(3)logaMn=nlogaM(n∈R)
(a & gt0，a≠ 1，M & gt0，N & gt0)
3. The relationship between exponential operation and logarithmic operation:
X = alogaxmlogan=nlogam
4. Negative numbers and zeros have no logarithm; The logarithm of 1 is zero, that is
loga 1 = 0； The logarithm of the base is 1, which means logaa= 1.
5. Logarithmic formula and its inference:
Bottom-changing formula: logaN=logbN/logba
Inference 1: logamnn = (n/m) Logan
Inference 2:
Third, exponential function and logarithmic function
Function y = ax(a & gt；; 0, and a≠ 1) is called exponential function. Its basic situation is:
The domains of (1) are all real numbers (-∞,+∞).
(2) The range is a positive real number (0, +∞), so the function has no maximum and minimum, only a lower bound, y >；; 0
(3) Correspondence is one-to-one mapping, so there is an inverse function-logarithmic function.
(4) Monotonicity is: when a >; Add functions in 1; When 0
(5) There is no parity, but the images of y=ax and y=a-x are symmetric about y axis, and the images of y=ax and y=-ax are symmetric about x axis; Images with y=ax and y=logax are symmetrical about the straight line y = x.
(6) There are two special points: zero point (0, 1) and invariant point (1, a).
(7) abstract nature: f (x) = ax (a >; 0，a≠ 1)，
f(x+y)=f(x) f(y)，f(x-y)=f(x)/f(y)
Function y = logax(a & gt；; 0, and a≠ 1) is called logarithmic function, and its basic situation is:
The domain of (1) is a positive real number (0, +∞).
(2) The range is all real numbers (-∞,+∞)
(3) The correspondence is one-to-one mapping, so there is an inverse function-exponential function.
(4) Monotonicity is: when a >; When 0, the increasing function at 1
(5) There is no parity. But y=logax and y=log( 1/a)x are symmetrical about x, y=logax and y=loga(-x) images are symmetrical about y, and y=logax and y=ax images are symmetrical about a straight line y = x.
(6) There are special points (1), (a), (a, 1)
(7) abstract operation property f (x) = logax (a >; 0，a≠ 1)，
f(x y)=f(x)+f(y)，
f(x/y)=f(x)-f(y)
Example 1. If f(x)=(ax/(ax+√a)), find f (11001)+f (2/1001)+f (3).
Analysis: There are 1000 items in the sum formula, so it is obviously not advisable to add them item by item. It is necessary to find out the structural characteristics of f(x) and find the law. Note that11001+1000/1= 2/10065438+999.
And f (x)+f (1-x) = (ax/(ax+√ a))+(a1-x/(a1-x+√ a)) = (ax/(ax+√ a).
Original formula = [f (11001)+f (1000/1)]+[f (2/1)
Note: Through observation and comparison, it is found that the rule of f(x)+f( 1-x)= 1 is a breakthrough in this example.
(1) take a=4, which is the fill-in-the-blank question of 1986 high school mathematics league: let f(x)=(4x/(4x+2)), then the formula f (11).
(2) If a=9 is selected, then f(x)=(9x/(9x+3)), and the summation formula can be changed to find f (1/n)+f (2/n)+f (3/n)+…+f ((n-).
(3) Let f(x)=( 1/(2x+√2)), and use the method of deducing the sum of the top n items of arithmetic progression in textbooks to get the value of f(-5)+f(-4)+…+f(0)+…+f(5)+f(6). This is the math 12 question of Shanghai college entrance examination in the spring of 2003.
Example 2.5log25 is equal to: ()
(A) 1/2 (B)( 1/5)
Solution: ∫ 5Log25 = (10/2) Log25 = (10Log25)/(2Log25) = (1/5) ×10Log25.
∴ Select (b)
Note: logarithmic identity is used here: A Logan = n (A > 0, A ≠ 1, n > 0).
This is the test question of Beijing 1997 Senior One Mathematics Competition.
Example 3. calculate
Solution 1: First, the true number is deformed by the collocation method of compound quadratic root simplification.
Solution 2: Using the basic properties of arithmetic roots to deform real numbers, there are
Description: multiplication formula can be used properly to simplify the complex.
Example 4. Compare (122002+1)/(122003+1) with (122003+1)/(1222)
Solution: For the size of two positive numbers, quotient is a common method to compare 1. If you record 122003 = a > 0, then there is.
(( 122002+ 1)/( 122003+ 1))； ( 122003+ 1)/( 122004+ 1)=((a/ 12)+ 1)/(a+65438
Therefore: ((122002+1)/(122003+1)) > ((122003+1).
Example 5. It is known that (a and b are real numbers) and f(lglog3 10)=5, then the value of f(lglg3) is ().
(A)-5 (B)-3 (C)3 (D) depends on the values of a and b.
Solution: let lglog3 10=t, then lglg3 = LG (1/log310) =-lglog310 =-T.
f(t)+f(-t)= 1
∴f(-t)=8-f(t)=8-5=3
Explanation: logab logba = (LGB/LGA) (LGA/LGB) =1,that is, logab=( 1/logba), so lglog3 10 and lglg3 are opposites. Set in part, then g(x) is odd function, and g(t)+g(-t)=0. This idea of holistic treatment skillfully uses odd function's properties to solve problems, and the key lies in carefully observing the structural characteristics of functions and the identity deformation of logarithms.
Example 6. Known function y = ((10x-10-x)/2) (x ∈ r)
(1) Find the inverse function y=f- 1(x)
(2) Judge whether the function y=f- 1(x) is an odd function or an even function.
Analysis: (1) Find the inverse function of y =( 10x- 10-x)/2. X is represented by y first, and then X and Y are reversed to get y = f- 1 (x).
(2) The parity of function y=f- 1(x) should be judged according to the definition of odd function or even function, and whether x ∈ r exists or not.
F(-x)=-f(x) or (f(-x)+f(x)=0)
Or f(-x)=f(x)
Keep building.
Solution: (1) From y = ((10x-10x)/2) (x ∈ r), we can get 2y = 10x- 10x.
Since t= 10x > 0, it is removed to obtain: substitute t= 10x into the above formula to obtain:
So the inverse function of the function y=(( 10x- 10-x)/2) is
(2) from:
∴f- 1(-x)=-f(x)
So, this function is odd function.
Description: ① A more general conclusion can be drawn from the process of solving and judging this problem: the inverse function of the function y = ((ax-a-x)/2) (x ∈ r, a > 0, a ≠ 1) is that they are all odd function. When a = 2,3, 10 or e, construct a new topic. Furthermore, we can study their monotonicity, such as the inverse function of the function y=((ex-e-x)/2), which is 1992.
(a) is odd function, which is a decreasing function at (0, +∞);
(b) is an even function, and is a decreasing function at (0, +∞);
(c) odd function, which is the increasing function on (0, +∞);
(d) is an even function and increasing function at (0, +∞).
② The function y=((ax-a-x)/2) is constructed from y=f(x)=ax, which is a full-time senior high school textbook (trial revision). Compulsory) Mathematics Volume I (1) (compiled by Middle School Mathematics Room of People's Education Publishing House) P 107 Review Reference Question 2 B Group Question 6: Let y=f(x) be an arbitrary function defined on R,
It is proved that (1) f1(x) = f (x)+f (-x) is an even function;
(2)F2(x)=f(x)-f(-x) is odd function.
And f(x)=F 1(X)+F2(x), which shows that any function defined on r can be expressed as the algebraic sum of a odd function (F2(x)) and an even function (F 1(X)). Starting from this proposition, many odd function can be constructed by f(x)=ax, such as y = ((ax-a-x)/2); Y = ((ax-a-x)/(ax+a-x)) = ((a2x-1)/(a2x+1)) and so on, the base of natural logarithm e ≈ 2.7 1828 ... (irrational
( 1)CH2(x)-sh2(x)= 1；
(2)sh(x+y)= sh(x)ch(y)+ch(x)sh(y)；
(3)ch(x+y)= ch(x)ch(y)+sh(x)sh(y)；
(4)th(x+y)=((th(x)+th(y))/( 1+th(x)th(y)))；
(5)ch(-x)= ch(x)；
(6)sh(-x)=-sh(x)；
(7)th(-x)=-th(x)。
Let x=y, then there is
(8)sh(2x)= 2sh(x)ch(x)；
ch(2x)=ch2(x)+sh2(x)
Among them, ① ⑧ ⑨ together is the eighth question in the textbook P 107.
Example 7. The function f (x) = loga ((1+x)/(1-x)) (a > 0, a ≠ 1) is known.
(1) Find the domain of f(x)
(2) Judge the parity of f(x) and prove it;
(3) when a > 1, find the value range of x that makes f (x) > 0;
(4) Find its inverse function f- 1(x).
Solution: (1) is known from the logarithm domain ((1+x)/( 1-x)) > 0.
Solve this fractional infinitive and get: (x+ 1) (x- 1) < 0,-1 < x < 1.
So the domain of the function f(x) is (-1, 1).
(2)f(-x)= loga(( 1-x)/( 1+x))= log(( 1+x)/( 1-x))- 1 =-loga(( 1+x)/( 1-x))=-f(x)
According to odd function's definition, the function f(x) is odd function.
(3)loga(( 1+x)/( 1-x))> 0 < = > loga(( 1+x)/( 1-x))> loga 1，
Because a > 1, we know from the monotonicity of logarithmic function ((1+x)/(1-x)) >1,and we know from (1)/
Therefore, for a > 1, when x∈(0, 1), the function f (x) > 0.
(4)from y = loga(( 1+x)/( 1-x)):(( 1+x)/( 1-x))= ay:((65438+)
∴x=((ay- 1)/(ay+ 1)) exchange x, y:
Y=((ax- 1)/(ax+ 1)), which is the inverse function of the function f (x) = loga ((1+x)/(kloc-0/-x).
Description: (1) function y = loga ((1+x)/(1-x)) and y = ((ax-1)) are a pair of inverse functions. Let a=e, and the domain of the inverse function of function y=((ex- 1)/(ex+ 1)) is. This is the topic of 1989 college entrance examination.
(2) f (x) = LG ((1-x)/(1+x)), a, B ∈ (- 1,1) Proof: F (a)+F (b)
(3)y=ax, y = logaxY=((ax-a-x)/2) and; Y=((ax- 1)/(ax+ 1)) and y = loga ((1+x)/(1-x)) and their properties need to be understood and remembered.
For example, the decimal representation of 8.22003 is p bit, and the decimal representation of 52003 is q bit, so p+q=.
Solution: ∫22003 is a p-digit number.
∴ 10p- 1 1②, so that LOGA (A2-A) < 0 = LOGA 1, A2-A.
Example 10. Let y = log (1/2) [a2x+2 (ab) x-b2x+1] (a > 0, b > 0) and find the value range of x that makes y negative.
Solution: ∫( 1/2)< 1, to make y < 0, just
a2x+2(ab)x-b2x+ 1> 1，
That is a2x+2 (ab) x-b2x > 0.
→b2x[(a/b)2x+2(a/b)x- 1]> 0
→[(a/b)x]2+2(a/b)x- 1>0
→
→∵
→.
1 when A > B > 0, A/B > 1,;
2 when b > a > 0, 0 < a/b < 1,
3 when a = b > 0, x ∈ r.
Exercise 4
1. Let a, b and c be positive numbers, and 3a=4b=6c, then ()
(A)( 1/C)=( 1/A)+( 1/B)，(B)(2/c)=(2/a)+( 1/b)，(C)( 1/c)=(2/a)+(2/b)，(D)(2/c)=( 1/a)+(2/b)
2. f(x) = (1+(2/(2x-1)) f(x) (x ≠ 0) is an even function, and f (x) is not always equal to zero, then f (x) ().
(a) is an odd function; (b) is an even function.
(c) It may be odd function or even function (d) It is not odd function or even function.
3. If f(x)=3x+5, the domain of f- 1(x) is ().
(A)(0，+∞) (B) (5，+∞) (C) (8，+∞) (D) (-∞，+∞)
4. Evaluation: 6lg40×5lg36
5. It is known that m and n are positive integers, a > 0, a ≠ 1, logam+loga (1+(1/m))+loga (1+(m)
6. The value of x = ((1/(log (1/2) (1/3))+(1/(log (1/5)).
(A)(-2，- 1) (B)( 1，2) (C)(-3，-2) (D)(2，3)
7. Calculation:
( 1)lg20+log 10025(2)lg5 lg20+(lg2)2
8. If the set {x, xy, lg(xy)}={0, ∣x∣,y}, then log8(x2+y2)=.
9. If x∈( 1, 10), the order of lg2x, lgx2 and lglgx is:
(A)lg2x < LG x2 < lglgx(B)lg2x < lglgx < LG x2
(C)LG x2 < lg2x < lglgx(D)lglglgx < lg2x < LG x2
10. Calculation:
1 1. The number of proper subset of the set {x ∣-1≤ log (1/x)10.
12. Find the monotone interval of function y=( 1/4)x2-2x-3.
13. Given the exponential function f (x) = ax (a > 0 and a≠ 1), find f (3x2-4x-5) >; The value of x of f (2x2-3x+ 1).
14. Solve equation 8log6(x2-7x+ 15)=5log68.
15. There is an inequality LG (∣ x+3 ∣+∣ x-7 ∣) > A.
(1) When a= 1, solve this inequality;
(2) When a is what value, the solution set of this inequality is r?
Reference answer
1.(B) and: 2. (1); 3.(B) and: 4.2 16; 5.m=2，n = 2；
6.(D) Providing information; 7.( 1)2,(2) 1; 8. 1/3; 9.(D) Providing information;
10. 1/2; 1 1.290- 1; 12. monotonically increasing interval (-∞, 1), monotonically decreasing interval [1, +∞)
13. when a > 1, x 3, when 0 < a < 1,-2 < x < 3;
14.x 1=2，x2 = 5；
15.( 1) x 7，(2) a < 1。