At the stage of preparation for the final test, high school students need to improve their knowledge on the topic “Exponential Equations.” The experience of past years indicates that such tasks cause certain difficulties for schoolchildren. Therefore, high school students, regardless of their level of preparation, need to thoroughly master the theory, remember the formulas and understand the principle of solving such equations. Having learned to cope with this type of problem, graduates can count on high scores when passing the Unified State Exam in mathematics.
Get ready for exam testing with Shkolkovo!
When reviewing the materials they have covered, many students are faced with the problem of finding the formulas needed to solve equations. A school textbook is not always at hand, and selecting the necessary information on a topic on the Internet takes a long time.
The Shkolkovo educational portal invites students to use our knowledge base. We are implementing a completely new method of preparing for the final test. By studying on our website, you will be able to identify gaps in knowledge and pay attention to those tasks that cause the most difficulty.
Shkolkovo teachers collected, systematized and presented all the material necessary for successfully passing the Unified State Exam in the simplest and most accessible form.
Basic definitions and formulas are presented in the “Theoretical background” section.
To better understand the material, we recommend that you practice completing the assignments. Carefully review the examples of exponential equations with solutions presented on this page to understand the calculation algorithm. After that, proceed to perform tasks in the “Directories” section. You can start with the easiest tasks or go straight to solving complex exponential equations with several unknowns or . The database of exercises on our website is constantly supplemented and updated.
Those examples with indicators that caused you difficulties can be added to “Favorites”. This way you can quickly find them and discuss the solution with your teacher.
To successfully pass the Unified State Exam, study on the Shkolkovo portal every day!
The free calculator we bring to your attention has a rich arsenal of possibilities for mathematical calculations. It allows you to use the online calculator in various fields of activity: educational, professional And commercial. Of course, using an online calculator is especially popular among students And schoolchildren, it makes it much easier for them to perform a variety of calculations.
At the same time, the calculator can become a useful tool in some areas of business and for people of different professions. Of course, the need to use a calculator in business or work is determined primarily by the type of activity itself. If your business and profession are associated with constant calculations and calculations, then it is worth trying out an electronic calculator and assessing the degree of its usefulness for a particular task.
This online calculator can
- Correctly perform standard mathematical functions written in one line like - 12*3-(7/2) and can process numbers larger than we can count huge numbers in an online calculator. We don’t even know what to call such a number correctly ( there are 34 characters and this is not the limit at all).
- Except tangent, cosine, sine and other standard functions - the calculator supports calculation operations arctangent, arccotangent and others.
- Available in Arsenal logarithms, factorials and other interesting features
- This online calculator knows how to build graphs!!!
To plot graphs, the service uses a special button (the graph is drawn in gray) or a letter representation of this function (Plot). To build a graph in an online calculator, just write the function: plot(tan(x)),x=-360..360.
We took the simplest graph for the tangent, and after the decimal point we indicated the range of the X variable from -360 to 360.
You can build absolutely any function, with any number of variables, for example this: plot(cos(x)/3z, x=-180..360,z=4) or even more complex that you can come up with. Pay attention to the behavior of the variable X - the interval from and to is indicated using two dots.
The only negative (although it is difficult to call it a disadvantage) of this online calculator is that it cannot build spheres and other three-dimensional figures - only planes.
How to use the Math Calculator
1. The display (calculator screen) displays the entered expression and the result of its calculation in ordinary symbols, as we write on paper. This field is simply for viewing the current transaction. The entry appears on the display as you type a mathematical expression in the input line.
2. The expression input field is intended for recording the expression that needs to be calculated. It should be noted here that the mathematical symbols used in computer programs are not always the same as those we usually use on paper. In the overview of each calculator function, you will find the correct designation for a specific operation and examples of calculations in the calculator. On this page below is a list of all possible operations in the calculator, also indicating their correct spelling.
3. Toolbar - these are calculator buttons that replace manual input of mathematical symbols indicating the corresponding operation. Some calculator buttons (additional functions, unit converter, solving matrices and equations, graphs) supplement the taskbar with new fields where data for a specific calculation is entered. The "History" field contains examples of writing mathematical expressions, as well as your six most recent entries.
Please note that when you press the buttons for calling additional functions, a unit converter, solving matrices and equations, and plotting graphs, the entire calculator panel moves up, covering part of the display. Fill in the required fields and press the "I" key (highlighted in red in the picture) to see the full-size display.
4. The numeric keypad contains numbers and arithmetic symbols. The "C" button deletes the entire entry in the expression entry field. To delete characters one by one, you need to use the arrow to the right of the input line.
Try to always close parentheses at the end of an expression. For most operations this is not critical; the online calculator will calculate everything correctly. However, in some cases errors may occur. For example, when raising to a fractional power, unclosed parentheses will cause the denominator of the fraction in the exponent to go into the denominator of the base. The closing bracket is shown in pale gray on the display and should be closed when recording is complete.
Key | Symbol | Operation |
---|---|---|
pi | pi | Constant pi |
e | e | Euler number |
% | % | Percent |
() | () | Open/Close Brackets |
, | , | Comma |
sin | sin(?) | Sine of angle |
cos | cos(?) | Cosine |
tan | tan(y) | Tangent |
sinh | sinh() | Hyperbolic sine |
cosh | cosh() | Hyperbolic cosine |
tanh | tanh() | Hyperbolic tangent |
sin -1 | asin() | Reverse sine |
cos -1 | acos() | Inverse cosine |
tan -1 | atan() | Reverse tangent |
sinh -1 | asinh() | Inverse hyperbolic sine |
cosh -1 | acosh() | Inverse hyperbolic cosine |
tanh -1 | atanh() | Inverse hyperbolic tangent |
x 2 | ^2 | Squaring |
x 3 | ^3 | Cube |
x y | ^ | Exponentiation |
10 x | 10^() | Exponentiation to base 10 |
e x | exp() | Exponentiation of Euler's number |
vx | sqrt(x) | Square root |
3 vx | sqrt3(x) | 3rd root |
yvx | sqrt(x,y) | Root extraction |
log 2 x | log2(x) | Binary logarithm |
log | log(x) | Decimal logarithm |
ln | ln(x) | Natural logarithm |
log y x | log(x,y) | Logarithm |
I/II | Minimize/Call additional functions | |
Unit | Unit converter | |
Matrix | Matrices | |
Solve | Equations and systems of equations | |
Graphing | ||
Additional functions (call with key II) | ||
mod | mod | Division with remainder |
! | ! | Factorial |
i/j | i/j | Imaginary unit |
Re | Re() | Isolating the whole real part |
Im | Im() | Excluding the real part |
|x| | abs() | The absolute value of a number |
Arg | arg() | Function argument |
nCr | ncr() | Binominal coefficient |
gcd | gcd() | GCD |
lcm | lcm() | NOC |
sum | sum() | Total value of all decisions |
fac | factorize() | Prime factorization |
diff | diff() | Differentiation |
Deg | Degrees | |
Rad | Radians |
In this video we will analyze a whole set of linear equations that are solved using the same algorithm - that’s why they are called the simplest.
First, let's define: what is a linear equation and which one is called the simplest?
A linear equation is one in which there is only one variable, and only to the first degree.
The simplest equation means the construction:
All other linear equations are reduced to the simplest using the algorithm:
- Expand parentheses, if any;
- Move terms containing a variable to one side of the equal sign, and terms without a variable to the other;
- Give similar terms to the left and right of the equal sign;
- Divide the resulting equation by the coefficient of the variable $x$.
Of course, this algorithm does not always help. The fact is that sometimes after all these machinations the coefficient of the variable $x$ turns out to be equal to zero. In this case, two options are possible:
- The equation has no solutions at all. For example, when something like $0\cdot x=8$ turns out, i.e. on the left is zero, and on the right is a number other than zero. In the video below we will look at several reasons why this situation is possible.
- The solution is all numbers. The only case when this is possible is when the equation has been reduced to the construction $0\cdot x=0$. It is quite logical that no matter what $x$ we substitute, it will still turn out “zero is equal to zero”, i.e. correct numerical equality.
Now let's see how all this works using real-life examples.
Examples of solving equations
Today we are dealing with linear equations, and only the simplest ones. In general, a linear equation means any equality that contains exactly one variable, and it goes only to the first degree.
Such constructions are solved in approximately the same way:
- First of all, you need to expand the parentheses, if there are any (as in our last example);
- Then combine similar
- Finally, isolate the variable, i.e. move everything connected with the variable—the terms in which it is contained—to one side, and move everything that remains without it to the other side.
Then, as a rule, you need to bring similar ones on each side of the resulting equality, and after that all that remains is to divide by the coefficient of “x”, and we will get the final answer.
In theory, this looks nice and simple, but in practice, even experienced high school students can make offensive mistakes in fairly simple linear equations. Typically, errors are made either when opening brackets or when calculating the “pluses” and “minuses”.
In addition, it happens that a linear equation has no solutions at all, or that the solution is the entire number line, i.e. any number. We will look at these subtleties in today's lesson. But we will start, as you already understood, with the simplest tasks.
Scheme for solving simple linear equations
First, let me once again write the entire scheme for solving the simplest linear equations:
- Expand the brackets, if any.
- We isolate the variables, i.e. We move everything that contains “X’s” to one side, and everything without “X’s” to the other.
- We present similar terms.
- We divide everything by the coefficient of “x”.
Of course, this scheme does not always work; there are certain subtleties and tricks in it, and now we will get to know them.
Solving real examples of simple linear equations
Task No. 1
The first step requires us to open the brackets. But they are not in this example, so we skip this step. In the second step we need to isolate the variables. Please note: we are talking only about individual terms. Let's write it down:
We present similar terms on the left and right, but this has already been done here. Therefore, we move on to the fourth step: divide by the coefficient:
\[\frac(6x)(6)=-\frac(72)(6)\]
So we got the answer.
Task No. 2
We can see the parentheses in this problem, so let's expand them:
Both on the left and on the right we see approximately the same design, but let's act according to the algorithm, i.e. separating the variables:
Here are some similar ones:
At what roots does this work? Answer: for any. Therefore, we can write that $x$ is any number.
Task No. 3
The third linear equation is more interesting:
\[\left(6-x \right)+\left(12+x \right)-\left(3-2x \right)=15\]
There are several brackets here, but they are not multiplied by anything, they are simply preceded by different signs. Let's break them down:
We perform the second step already known to us:
\[-x+x+2x=15-6-12+3\]
Let's do the math:
We carry out the last step - divide everything by the coefficient of “x”:
\[\frac(2x)(x)=\frac(0)(2)\]
Things to Remember When Solving Linear Equations
If we ignore too simple tasks, I would like to say the following:
- As I said above, not every linear equation has a solution - sometimes there are simply no roots;
- Even if there are roots, there may be zero among them - there is nothing wrong with that.
Zero is the same number as the others; you shouldn’t discriminate against it in any way or assume that if you get zero, then you did something wrong.
Another feature is related to the opening of brackets. Please note: when there is a “minus” in front of them, we remove it, but in parentheses we change the signs to opposite. And then we can open it using standard algorithms: we will get what we saw in the calculations above.
Understanding this simple fact will help you avoid making stupid and hurtful mistakes in high school, when doing such things is taken for granted.
Solving complex linear equations
Let's move on to more complex equations. Now the constructions will become more complex and when performing various transformations a quadratic function will appear. However, we should not be afraid of this, because if, according to the author’s plan, we are solving a linear equation, then during the transformation process all monomials containing a quadratic function will certainly cancel.
Example No. 1
Obviously, the first step is to open the brackets. Let's do this very carefully:
Now let's take a look at privacy:
\[-x+6((x)^(2))-6((x)^(2))+x=-12\]
Here are some similar ones:
Obviously, this equation has no solutions, so we’ll write this in the answer:
\[\varnothing\]
or there are no roots.
Example No. 2
We perform the same actions. First step:
Let's move everything with a variable to the left, and without it - to the right:
Here are some similar ones:
Obviously, this linear equation has no solution, so we’ll write it this way:
\[\varnothing\],
or there are no roots.
Nuances of the solution
Both equations are completely solved. Using these two expressions as an example, we were once again convinced that even in the simplest linear equations, everything may not be so simple: there can be either one, or none, or infinitely many roots. In our case, we considered two equations, both simply have no roots.
But I would like to draw your attention to another fact: how to work with parentheses and how to open them if there is a minus sign in front of them. Consider this expression:
Before opening, you need to multiply everything by “X”. Please note: multiplies each individual term. Inside there are two terms - respectively, two terms and multiplied.
And only after these seemingly elementary, but very important and dangerous transformations have been completed, can you open the bracket from the point of view of the fact that there is a minus sign after it. Yes, yes: only now, when the transformations are completed, we remember that there is a minus sign in front of the brackets, which means that everything below simply changes signs. At the same time, the brackets themselves disappear and, most importantly, the front “minus” also disappears.
We do the same with the second equation:
It is not by chance that I pay attention to these small, seemingly insignificant facts. Because solving equations is always a sequence of elementary transformations, where the inability to clearly and competently perform simple actions leads to the fact that high school students come to me and again learn to solve such simple equations.
Of course, the day will come when you will hone these skills to the point of automaticity. You will no longer have to perform so many transformations each time; you will write everything on one line. But while you are just learning, you need to write each action separately.
Solving even more complex linear equations
What we are going to solve now can hardly be called the simplest task, but the meaning remains the same.
Task No. 1
\[\left(7x+1 \right)\left(3x-1 \right)-21((x)^(2))=3\]
Let's multiply all the elements in the first part:
Let's do some privacy:
Here are some similar ones:
Let's complete the last step:
\[\frac(-4x)(4)=\frac(4)(-4)\]
Here is our final answer. And, despite the fact that in the process of solving we had coefficients with a quadratic function, they canceled each other out, which makes the equation linear and not quadratic.
Task No. 2
\[\left(1-4x \right)\left(1-3x \right)=6x\left(2x-1 \right)\]
Let's carefully perform the first step: multiply each element from the first bracket by each element from the second. There should be a total of four new terms after the transformations:
Now let’s carefully perform the multiplication in each term:
Let’s move the terms with “X” to the left, and those without - to the right:
\[-3x-4x+12((x)^(2))-12((x)^(2))+6x=-1\]
Here are similar terms:
Once again we have received the final answer.
Nuances of the solution
The most important note about these two equations is the following: as soon as we begin to multiply brackets that contain more than one term, this is done according to the following rule: we take the first term from the first and multiply with each element from the second; then we take the second element from the first and similarly multiply with each element from the second. As a result, we will have four terms.
About the algebraic sum
With this last example, I would like to remind students what an algebraic sum is. In classical mathematics, by $1-7$ we mean a simple construction: subtract seven from one. In algebra, we mean the following by this: to the number “one” we add another number, namely “minus seven”. This is how an algebraic sum differs from an ordinary arithmetic sum.
As soon as, when performing all the transformations, each addition and multiplication, you begin to see constructions similar to those described above, you simply will not have any problems in algebra when working with polynomials and equations.
Finally, let's look at a couple more examples that will be even more complex than the ones we just looked at, and to solve them we will have to slightly expand our standard algorithm.
Solving equations with fractions
To solve such tasks, we will have to add one more step to our algorithm. But first, let me remind you of our algorithm:
- Open the brackets.
- Separate variables.
- Bring similar ones.
- Divide by the ratio.
Alas, this wonderful algorithm, for all its effectiveness, turns out to be not entirely appropriate when we have fractions in front of us. And in what we will see below, we have a fraction on both the left and the right in both equations.
How to work in this case? Yes, it's very simple! To do this, you need to add one more step to the algorithm, which can be done both before and after the first action, namely, getting rid of fractions. So the algorithm will be as follows:
- Get rid of fractions.
- Open the brackets.
- Separate variables.
- Bring similar ones.
- Divide by the ratio.
What does it mean to “get rid of fractions”? And why can this be done both after and before the first standard step? In fact, in our case, all fractions are numerical in their denominator, i.e. Everywhere the denominator is just a number. Therefore, if we multiply both sides of the equation by this number, we will get rid of fractions.
Example No. 1
\[\frac(\left(2x+1 \right)\left(2x-3 \right))(4)=((x)^(2))-1\]
Let's get rid of the fractions in this equation:
\[\frac(\left(2x+1 \right)\left(2x-3 \right)\cdot 4)(4)=\left(((x)^(2))-1 \right)\cdot 4\]
Please note: everything is multiplied by “four” once, i.e. just because you have two parentheses doesn't mean you have to multiply each one by "four." Let's write down:
\[\left(2x+1 \right)\left(2x-3 \right)=\left(((x)^(2))-1 \right)\cdot 4\]
Now let's expand:
We seclude the variable:
We perform the reduction of similar terms:
\[-4x=-1\left| :\left(-4 \right) \right.\]
\[\frac(-4x)(-4)=\frac(-1)(-4)\]
We have received the final solution, let's move on to the second equation.
Example No. 2
\[\frac(\left(1-x \right)\left(1+5x \right))(5)+((x)^(2))=1\]
Here we perform all the same actions:
\[\frac(\left(1-x \right)\left(1+5x \right)\cdot 5)(5)+((x)^(2))\cdot 5=5\]
\[\frac(4x)(4)=\frac(4)(4)\]
The problem is solved.
That, in fact, is all I wanted to tell you today.
Key points
Key findings are:
- Know the algorithm for solving linear equations.
- Ability to open brackets.
- Don't worry if you have quadratic functions somewhere; most likely, they will be reduced in the process of further transformations.
- There are three types of roots in linear equations, even the simplest ones: one single root, the entire number line is a root, and no roots at all.
I hope this lesson will help you master a simple, but very important topic for further understanding of all mathematics. If something is not clear, go to the site and solve the examples presented there. Stay tuned, many more interesting things await you!
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