Que bién, pero se convierte en larga explicación ,hasta llega a usar Logaritmos ,complejo para mi entendimiento matemático , saludos desde Nicaragua , Felicidades
Why did you not mention that with common logs one cannot use a solution that is complex? If one tries to substitute the single complex value, there is no way to get a real value back out of the original expression.
Maybe this way is simpler: let 7^(x)=a a+1+a²+a³=15 (a+1)+a²(a+1)=15 (a+1)(a²+1)=15=5x3 °°°Case 1 (a+1)=5 and (a²+1)= 3 There is no satisfactory a solution °°°Case 2 (a+1)=3 and (a²+1)= 5 So a=2
One doesn't need a pen, let alone the 15 min explanation: y+y²+y³=14 => y=2 (easily seen). Besides, 2 is the only solution (not considering complex numbers) as y+y²+y³ is monotonuos. Hence x = 1/log7 (base 2)
7ˣ+7²ˣ+7³ˣ=14 7ˣ+(7ˣ⨯7ˣ)+(7ˣ⨯7ˣ⨯7ˣ)=14 PUT 7ˣ=X X+X²+X³=14 X³+X²+X-14=0 It is third degree equation Its solutions: X=2 Or X=((i√19)-3)/2 } By calculator Or X=((-i√19)-3)/2 So: X=2 For X=7ˣ positive real number So: 7ˣ=2 For 7>2 ,must x
7^x+7^2x+7^3x=14 Or 7^x+7^3x=14-7^2x Or 7^x(1+7^2x) =14--7^2x Or log 7^x+log7^3x-log14- -log7^2x=0 Or 4log7^x-2log7^x-log7-log2=0 Or 2log7^x-log7-log2=0 Or 2xlog7=log7+log2 Or x=(log7+log2) /2log7 =1/2+(log2/2log7) Or 2.49^x=13 Or 49^x=13/2 Or xlog49=log13-log2 Or x=(log13-log2) /2log7 Or 7
Why didn't you verify the complex results ... because it is a trivial solution and you didn't need it ... once the second expression gives a comples solution you should have dumbed it and enough with the real value in the 1st one
How can you take 14 minutes, when I did it in my head in less than a minute y=7^x Y+y^2 +y^3 =14 y=2 works 2+2^2+2^3 = 2+4+8=14 LHS is monotonic increasing, rhs is constant so there is only one real solution 2=7^x Log2/log7=x
Hola buenos días como estas todo ustedes desde San Felipe de puerto plata primera espalda de la restauración general gregorio Luperón machete carajo 🇩🇴🍫🍫☕
To solve the equation: \[ 7^{x} + 7^{2x} + 7^{3x} = 14 \] we need to find the real value of \( x \) that satisfies this equation. Step 1: Simplify the Equation Let's let \( y = 7^{x} \). Then the equation becomes: \[ y + y^{2} + y^{3} = 14 \] Step 2: Form a Polynomial Equation We can rewrite the equation as: \[ y^{3} + y^{2} + y - 14 = 0 \] Step 3: Factor the Polynomial We can attempt to factor the cubic polynomial. Let's test possible rational roots using the Rational Root Theorem. The possible rational roots are factors of the constant term \( -14 \) divided by factors of the leading coefficient \( 1 \): Possible roots: \( \pm1, \pm2, \pm7, \pm14 \) Let's test \( y = 2 \): \[ (2)^{3} + (2)^{2} + 2 - 14 = 8 + 4 + 2 - 14 = 0 \] Since \( y = 2 \) satisfies the equation, it is a root. We can factor out \( (y - 2) \): \[ (y - 2)(y^{2} + 3y + 7) = 0 \] Now, we have: \( y - 2 = 0 \) ⇒ \( y = 2 \) \( y^{2} + 3y + 7 = 0 \) The quadratic equation \( y^{2} + 3y + 7 = 0 \) has a discriminant: \[ D = b^{2} - 4ac = (3)^{2} - 4(1)(7) = 9 - 28 = -19 \] Since the discriminant is negative, the quadratic equation has no real solutions. Therefore, the only real solution is \( y = 2 \). Step 4: Solve for \( x \) Recall that \( y = 7^{x} \), so: \[ 7^{x} = 2 \] To solve for \( x \), take the logarithm of both sides. Using the natural logarithm (ln) or logarithm base 7: Using Natural Logarithm: \[ x \ln 7 = \ln 2 \] \[ x = \frac{\ln 2}{\ln 7} \] Using Logarithm Base 7: \[ x = \log_{7} 2 \] Step 5: Calculate the Approximate Value Using a calculator to compute the natural logarithms: \[ x = \frac{\ln 2}{\ln 7} \approx \frac{0.6931}{1.9459} \approx 0.3562 \] Answer: The value of \( x \) that satisfies the equation is: \[ x = \log_{7} 2 \approx 0.3562 \]
It’s the first time that I see too many diamonds rings in the fingers of a woman solving equations…women are powerful…I can wear not even one ring on my finger, it bothers me all day long, not to mention too many…
As a lover of mathematics, this explanation is remarkable ❤
As a lover of Mathematics, this explanation is exceptional!!!!!
Wow. Good teaching
Que bién, pero se convierte en larga explicación ,hasta llega a usar Logaritmos ,complejo para mi entendimiento matemático , saludos desde Nicaragua , Felicidades
liked it very much
You tell it in a very nice and cheerful way. You are wonderful. I wish you success in your career.❤️
Thank you so much!
Why did you not mention that with common logs one cannot use a solution that is complex? If one tries to substitute the single complex value, there is no way to get a real value back out of the original expression.
That's the point I thought about .. if I took "ln" function for both sides it will be easier to solve
Effective teaching
7^x=a; a + a^2 + a^3 = 14; a(1+a+a^2) =14 ; a= 2 ; 1+a+a^2=7 ; 7^x =2 ; I have written before seeing the video, unfortunately the same😂
I believe you wrote it before seeing the video, because your answer is wrong. The question is to find X and you haven’t done so.
@jasonvuong2656 rest of the solution is for kids. Ok then, 7^X = 2; X = Log base 7 of 2
Perfect example to watch
Add +1 to both the sides
(7^x+1)(7^2x+1)=15
Now equate 7^x+1=3 and 7^2x+1=5
Good teaching method
ruclips.net/video/oz_yxjkJ5VY/видео.html
Maybe this way is simpler: let 7^(x)=a
a+1+a²+a³=15
(a+1)+a²(a+1)=15
(a+1)(a²+1)=15=5x3
°°°Case 1 (a+1)=5 and (a²+1)= 3
There is no satisfactory a solution
°°°Case 2 (a+1)=3 and (a²+1)= 5
So a=2
@shaikh0087 Why can't i add 1?
(a+1)=5 then a=4 but (a²+1)=3 then a= sqrt(2)
So this case is not satisfied
One doesn't need a pen, let alone the 15 min explanation:
y+y²+y³=14 => y=2 (easily seen).
Besides, 2 is the only solution (not considering complex numbers) as y+y²+y³ is monotonuos.
Hence x = 1/log7 (base 2)
Can we solve by log method I thought it will be less easy.
Your method is effective too. 👍
7^x+7^2x+7^3x=14=7^1+7^1-->x+2x+3x=2-->×=1/3 😅 intuition method😅
Wrong solution, x=log2/log7
Nice explanation
It could have been easy to understand by students if she used the short concept.
Đặt y=7^x được phương trình :
y^3+y^2+y = 14
well explained
7ˣ+7²ˣ+7³ˣ=14
7ˣ+(7ˣ⨯7ˣ)+(7ˣ⨯7ˣ⨯7ˣ)=14
PUT 7ˣ=X
X+X²+X³=14
X³+X²+X-14=0
It is third degree equation
Its solutions:
X=2
Or
X=((i√19)-3)/2 } By calculator
Or
X=((-i√19)-3)/2
So:
X=2
For X=7ˣ positive real number
So:
7ˣ=2
For 7>2 ,must x
Excellence
Log 2 base 7
God bless you
Well ma'am you teach really good but I want to know to solve the same in complex value which you got can you please elaborate also
Yes definitely 😊
X= log2/log7 (log2 with base7)
X=1
Respect
Or have a computer solve it by iteration. In which case: x = 0.3562 (positive root only solution)
NICE 🎉NAMASTE 1:56 1:57
7^x+7^2x+7^3x=14
Or 7^x+7^3x=14-7^2x
Or 7^x(1+7^2x) =14--7^2x
Or log 7^x+log7^3x-log14-
-log7^2x=0
Or 4log7^x-2log7^x-log7-log2=0
Or 2log7^x-log7-log2=0
Or 2xlog7=log7+log2
Or x=(log7+log2) /2log7
=1/2+(log2/2log7)
Or 2.49^x=13
Or 49^x=13/2
Or xlog49=log13-log2
Or x=(log13-log2) /2log7
Or 7
I’ve a problem with x3. The log of a negative number does not exist
lovely ji
我用Rational Zero Theorem去求m^3+m^2+m-14=0
所以很快得到(m-2)(m^2+3m+7)兩個方程
後面的步驟大致上都跟影片差不多
Why didn't you verify the complex results ... because it is a trivial solution and you didn't need it ... once the second expression gives a comples solution you should have dumbed it and enough with the real value in the 1st one
I eish i had a math teacher like u
0.3562
गुड टीचिंग
Log2/log7 = log 2 to the base 7 is correct???!
How can you take 14 minutes, when I did it in my head in less than a minute
y=7^x
Y+y^2 +y^3 =14
y=2 works
2+2^2+2^3 = 2+4+8=14
LHS is monotonic increasing, rhs is constant so there is only one real solution
2=7^x
Log2/log7=x
How did log2/log7=log2,base 7?
= 3
Why you did not use common logs ???
Pehle hi take log to base 7 on both sides.
That means X is very small!!!😁😊😁
Could you prove that the imaginary result also solution for the equation ... ??? ... 😂😂😂
Racine carré de(-1) i est égale à la valeur absolue de i et non pas i
Bu sorular turkiyede lise seviyesi matematik sorularidir
Why are you using such lengthy method
I did it a different way
Io mi sarei vergognata di svolgere l'esercizio in questo modo....ma chi è questo prof che scrive?😢
Hola buenos días como estas todo ustedes desde San Felipe de puerto plata primera espalda de la restauración general gregorio Luperón machete carajo 🇩🇴🍫🍫☕
И это олимпиадное? Да это же ЕГЭ номер 7 (почти)
Complex logarithme is not a function. You can not use it as function !!!
make long
Horner
Taking 30 minutes working for 3 marks. Not me
7x = t
I try to do this calculation by other method.
Incorrect the racine doivent positif
Cok aşırı uzatmış. Ben otobüste zihinden çözdüm
To solve the equation:
\[ 7^{x} + 7^{2x} + 7^{3x} = 14 \]
we need to find the real value of \( x \) that satisfies this equation.
Step 1: Simplify the Equation
Let's let \( y = 7^{x} \). Then the equation becomes:
\[ y + y^{2} + y^{3} = 14 \]
Step 2: Form a Polynomial Equation
We can rewrite the equation as:
\[ y^{3} + y^{2} + y - 14 = 0 \]
Step 3: Factor the Polynomial
We can attempt to factor the cubic polynomial. Let's test possible rational roots using the Rational Root Theorem. The possible rational roots are factors of the constant term \( -14 \) divided by factors of the leading coefficient \( 1 \):
Possible roots: \( \pm1, \pm2, \pm7, \pm14 \)
Let's test \( y = 2 \):
\[ (2)^{3} + (2)^{2} + 2 - 14 = 8 + 4 + 2 - 14 = 0 \]
Since \( y = 2 \) satisfies the equation, it is a root. We can factor out \( (y - 2) \):
\[ (y - 2)(y^{2} + 3y + 7) = 0 \]
Now, we have:
\( y - 2 = 0 \) ⇒ \( y = 2 \)
\( y^{2} + 3y + 7 = 0 \)
The quadratic equation \( y^{2} + 3y + 7 = 0 \) has a discriminant:
\[ D = b^{2} - 4ac = (3)^{2} - 4(1)(7) = 9 - 28 = -19 \]
Since the discriminant is negative, the quadratic equation has no real solutions. Therefore, the only real solution is \( y = 2 \).
Step 4: Solve for \( x \)
Recall that \( y = 7^{x} \), so:
\[ 7^{x} = 2 \]
To solve for \( x \), take the logarithm of both sides. Using the natural logarithm (ln) or logarithm base 7:
Using Natural Logarithm:
\[ x \ln 7 = \ln 2 \]
\[ x = \frac{\ln 2}{\ln 7} \]
Using Logarithm Base 7:
\[ x = \log_{7} 2 \]
Step 5: Calculate the Approximate Value
Using a calculator to compute the natural logarithms:
\[ x = \frac{\ln 2}{\ln 7} \approx \frac{0.6931}{1.9459} \approx 0.3562 \]
Answer:
The value of \( x \) that satisfies the equation is:
\[ x = \log_{7} 2 \approx 0.3562 \]
What a tedious work
Long winded way of getting answers…
I can't understand the
It’s the first time that I see too many diamonds rings in the fingers of a woman solving equations…women are powerful…I can wear not even one ring on my finger, it bothers me all day long, not to mention too many…
7^x=2 and x=log2/log7 No need to watch
Find the x. Pride + x = 2pride 😂😂😂😂😂
I' m not stupide
If you first took log on both side no need to do elaborate part
Pagal hai ye.... solve karte karte kaha nikal gayi...
कि सका किससे गुणा कर रही स्पष्ट नही ।
पूरी गलत