How to express the log function based on 3 in MATLAB

How to express the log function based on 3 in MATLAB

Matlab defines log2, log10 and E as the base of the log, the rest of the number of the base is not defined, but you can get the logarithm of any integer according to the bottom formula. The bottom formula: logx (y) = log (y) / log (x) will be converted from X as the base to e as the base

The matlab program of decomposing 1 * 1000 vector into 20 * 50 matrix

1000 is too much, use 10
>> A=1:10
A =
1 2 3 4 5 6 7 8 9 10
>> B=reshape(A,2,5)
B =
1 3 5 7 9
2 4 6 8 10

8*2=8+9=17 5*3=5+6+7=18 4*=4+5+6+7+8+9=39 10*2（ ）

Guess 10 * 2 = 10 + 11 = 21
Is there a number missing in your question? 4 *? = 4 + 5 + 6 + 7 + 8 + 9 = 39

How to solve the equation: 1 / 5x + 450 = 3 / 10x

1/5x+450=3/10x
1/5x+450=1/5x+1/10x
450=1/10x
x=1/4500

Is the distance from a point on an ellipse to two focal points equal to the major axis?

We are studying the standard equation of ellipse. You are talking about his definition. This point is the two vertices of the minor axis

Sequence 1, root 5, 3, root 13, root 17 What is the general term formula of

4n-3 under radical

0.5 three minus x plus 0.2 x plus four is greater than or equal to 14

After simplification. 5 / 30-x + 2 / 10x ≥ 14-4
60-10x = 50x ≥ 10x10
60=40x≥100
40x≥40
x≥1

How many parts of a meter is 79 decimeters?

One meter equals one decimeter, so 79 decimeter is 79 / 10, so you are right

1/(1×3×5) +1/(3×5×7)+1/(5×7×9)+… +How to use the formula of 1 / (11 × 13 × 15)

1/(1×3×5) +1/(3×5×7)+1/(5×7×9)+… +1/(11×13×15)=1/2x[1/1x3-1/3x5]+1/2[1/3x5-1/5x7]+.+1/2[1/11x13-1/13x15]=1/2[1/1x3-1/3x5+1/3x5-1/5x7+.+1/11x13-1/13x15]=1/2x[1/1x3-1/13x15]=1/2x64/195=32/195

Several solutions of Tree Science 3-variable quadratic equation

An example of solving the system of linear equations with three variables
Objectives and requirements
1. Understand the concept of ternary linear equations; master the simple solution of ternary linear equations; be able to choose simple and special solution to solve special ternary linear equations
2. Through the training of using substitution elimination method, addition and subtraction elimination method to solve simple ternary linear equations and the selection of reasonable and simple method to solve equations, cultivate the ability of operation
3. Through the observation and analysis of the characteristics of the coefficients of unknowns in the equations, it is clear that the main idea of solving the ternary linear equations is
In order to promote the transformation from unknown to known, cultivate and develop logical thinking ability
4. Through the study of the method of transforming the ternary linear equations into binary linear equations after elimination, and then transforming the elimination into unitary linear equations, and transforming some algebraic problems into equations, we can cultivate the ability of solving problems and developing thinking ability
[knowledge points]
1. The concept of linear equations of three variables
There are three unknowns, the degree of the unknown term of each equation is 1, and there are three equations in total
For example:
They are called systems of linear equations of three variables
Note: each equation does not necessarily contain three unknowns, but the equations as a whole contain three unknowns
Master the simple solution of ternary linear equations
Will describe the simple solution of ternary linear equations ideas and steps
Idea: the basic idea of solving linear equations of three variables is elimination, and its basic methods are substitution and addition and subtraction
The steps are as follows: (1) using the substitution method or addition and subtraction method, an unknown number is eliminated, and a binary linear equation system is obtained;
② By solving this system of linear equations of two variables, the values of two unknowns are obtained;
③ The values of the two unknowns are substituted into the simpler one of the original equations, and the values of the third unknowns are obtained
These three numbers together are the solutions of the system of linear equations of three variables
Flexible use of addition and subtraction elimination method, substituting elimination method to solve simple ternary linear equations
For example, solve the following system of linear equations of three variables
Analysis: this system of equations can be substituted by the substitution method to eliminate y first, and substitute ① into ② to get,
5x+3(2x-7)+2z=2
5x+6x-21+2z=2
By solving the linear equations of two variables, we get the following results
Substituting x = 2 into (1) yields y = - 3
Example 2
Analysis: the solution of ternary linear equations is similar to the solution of binary linear equations. When eliminating elements, it is better to choose the unknowns with simple coefficients. In the above ternary linear equations, considering the coefficient characteristics of the unknowns of the three equations, it is easier to eliminate Z first
Solution: 1 + 2, 5x + y = 26, 4
① In addition, 3x + 5Y = 42
④ And (5) constitute the equations
To solve this system of equations, we have to
Substituting it into equation 3, which is convenient for calculation, we get z = 8
∴
Note: in order to transform the ternary system of linear equations into binary system of linear equations, each equation in the original system should be used at least once
Can choose simple, special solution to solve special ternary linear equations
For example, solve the following system of linear equations of three variables
Analysis: in this system of equations, x, y, Z have the same number of times and the same coefficient
It is easy to solve the problem by adding the two sides separately
Solution: 1 + 2 + 3: 2 (x + y + Z) = 30
x+y+z=15④
Then, we get z = 5
④ (2) y = 9
④ - x = 1
∴
Analysis: according to the characteristics of the equations, equations 1 and 2 give the proportional relationship. First, let x = 3k, y = 2K, from which we can get z = y, z = × 2K = k, and then substitute x = 3k, y = 2K, z = K into 3, we can get the value of K, and then get the value of X, y, Z
Solution: let x = 3k, y = 2K
Let z = y = × 2K = K
Substituting x = 3k, y = 2K, z = K into 3, we get
3K + 2K + k = 66, k = 10
∴x=3k=30
y=2k=20
z=k=16