@@danishasfa1677 It is obviously that 1 is one of the roots. So, the product of roots is equat to 1 times another root. This implies another root is equal to the product of roots of the quadratic equation. 😏😏😏
More simply: divide both sides by 3*7 3^(x-1) * 7^(x²-1) = 1 Since x²-1 = (x-1)*(x+1): 3^(x-1) * (7^(x+1))^(x-1) = 1 Either x-1=0 (=> x=1) or we can take the (x-1) root on both side: 3 * 7^(x+1) = 1 Divide again by 3*7: 7^x = 1/21 and taking the log gives us the other solution.
@@copernic7511 You say that x=1 is a solution right? But it's wrong! Substitute X for 1 in the main equation to prove it! The main equation is 3^x * 7^x^2 = 21 . The main equation is not 3^x * 7^(x^2) = 21
For simplification the calculations let's designate the constant a = ln 3 / ln 7 Logarithm of both sides of the equation a•x + x²=1+a The first root is quite obvious x=1 The second root can be found with the Viete theorem: x1•x2 = - (1+a) So x = - (1+a) = - (1 + ln 3/ln 7)
It's true that you can easily spot one solution, when x = 1, since 3 * 7 = 21. Knowing that makes it easy at 3:05 to take a factor of (x-1) out of the quadratic (log7)x^2 + (log3)x - log21 = 0. That gives you (x-1) * (x*log7 + log21) = 0. So the other solution will be when x*log7 + log21 = 0, which gives x = - log21/log7 or - 1 - log3/log7 .
@@LAL666666 The problem is that you've confused 7^x^2 with 7^2x. If x=1 then x^2 = 1, not 2. So 7^x^2 = 7^1 = 7. Multiply that by 3 and you get 21. When you evaluate a "tower" of exponents, you always evaluate from the top downwards.
Can be done by examining the factors of 21 and then solving all the equations? Like the first case with 3^x=3 and 7^x^2=7 and second case with 21 and 1 and then checking the solutions
hmm, I got a different approach. I used logarithms as well though. ln[3^x * 7^(x^2)] = ln(21) --> ln(3^x) + ln(7^(x^2)) = ln(21) --> x * ln(3) + x^2 * ln(7) = ln(21) = ln(3 * 7) = ln(3) + ln(7). Put everything on the RHS to the left and group: x * ln(3) - ln(3) + x^2 * ln(7) - ln(7) = 0 --> (x - 1) * ln(3) + (x^2 - 1) * ln(7) = 0 --> (x - 1) * ln(3) + (x - 1)(x + 1) * ln(7) = 0 --> (x - 1)[ln(3) + (x + 1) * ln(7)] = 0. So either x - 1 = 0 or ln(3) + (x + 1) * ln(7) = 0. So the first one gives x = 1 and the second one gives (x + 1) * ln(7) = -ln(3) = ln(1/3), so x + 1 = ln(1/3) / ln(7) = log_7(1/3) --> x = log_7(1/3) - 1 = log_7(1/3) - log_7(7) = log_7(1/21). This last one is the same as -ln(21)/ln(7), but I just find it a neater answer like this.
@@LAL666666 You say that 7^x^2 = (7^x)^2. I don't think that's correct. To my knowledge 7^x^2 is indeed 7^x² because if you don't specify the order of exponentiation with brackets you'd start on the right. That is a^b^c^d = a^(b^(c^d)) not a^b^c^d = a^(bcd).
I really like your approach. Despite your name, it's pretty cool and no Quatsch at all! 😛 Personally, I wouldn't turn the logarithmic arguments into fractions or change the bases, at least not until the end, but that's of course a personal preference. You could then proceed as follows: x + 1 = -ln(3) / ln(7) x = -ln(3) / ln(7) - 1 = -ln(3) / ln(7) - ln(7) / ln(7) = [-ln(3) - ln(7)] / ln(7) = -ln(21) / ln(7) = -ln₇(21) Again, really cool approach!
So SyberMaths, For many of the final equation results you present, please do complete the arithmetic to show the value of the results. So here: -ln21/ln7 = ??? I then need my calculator to find the numerator and the denominator to then show the result, to compare with 1. Do this for all such results, since you are much more fluid than many of us are. Thank you 🤓 Cheers - - -
Señores SyberMath. Reciban un cordial saludo, Gracias por este lindo ejercicio, GRACIAS por compartir esa visión que nos permite tener herramientas para afrontar un problema. Felicidades, como la que tengo YO después de ver este desarrollo.
Believe it or not, I solved it exactly the same way. First, I established the quadratic equation using the ln(): ln(7)*x^2 + ln(3)*x - ln(21) = 0 Then divided by ln(7): x^2 + ln(3)/ln(7)*x - ln(21)/ln(7) = 0 First, I tried to simplify the discriminant and quadratic formula, but found it too complicatd. Then I remembered that by plugging in, x = 1 is the first solution x1, and I can use Vieta's laws: x1 * x2 = -ln(21)/ln(7) So since x1 = 1, we get x2 = - ln(21)/ln(7) = - log7(21) = -1.564575...
3^x = 21 / 7^x^2 3^x = 3/1^x^2 Because 1^x^2 always 1 in the number and negative number are also going to positive like -1² is 1, -2² is 2 and either, so we can make the solution at below that 1^x^2 is 1 so 3^x = 3 And we can conclude that x = 1 Is it true or not?(Sorry for my bad English)
At the end, using the product of roots to find another root is much more faster and easier. Very nice!!! 😊😊😊😊😊😊
Yes, you are right
Its so easy; it almost feels like cheating.
how?
@@danishasfa1677 It is obviously that 1 is one of the roots. So, the product of roots is equat to 1 times another root. This implies another root is equal to the product of roots of the quadratic equation. 😏😏😏
More simply: divide both sides by 3*7
3^(x-1) * 7^(x²-1) = 1
Since x²-1 = (x-1)*(x+1):
3^(x-1) * (7^(x+1))^(x-1) = 1
Either x-1=0 (=> x=1) or we can take the (x-1) root on both side:
3 * 7^(x+1) = 1
Divide again by 3*7:
7^x = 1/21
and taking the log gives us the other solution.
Nice
@@SyberMath no, no, no!
a*b=1 don´t mean that a=1 and b=1
@@LAL666666 It's not about that. I just jumped over the step ( 3 * 7^(x+1) )^(x-1) = 1 and that's where we can separate the cases x-1=0 and x-1≠0.
@@copernic7511 You say that x=1 is a solution right?
But it's wrong!
Substitute X for 1 in the main equation to prove it!
The main equation is 3^x * 7^x^2 = 21 .
The main equation is not 3^x * 7^(x^2) = 21
For simplification the calculations let's designate the constant a = ln 3 / ln 7
Logarithm of both sides of the equation
a•x + x²=1+a
The first root is quite obvious x=1
The second root can be found with the Viete theorem: x1•x2 = - (1+a)
So x = - (1+a) = - (1 + ln 3/ln 7)
It's true that you can easily spot one solution, when x = 1, since 3 * 7 = 21. Knowing that makes it easy at 3:05 to take a factor of (x-1) out of the quadratic (log7)x^2 + (log3)x - log21 = 0.
That gives you (x-1) * (x*log7 + log21) = 0. So the other solution will be when x*log7 + log21 = 0, which gives x = - log21/log7 or - 1 - log3/log7 .
if x=1 you have 3 * 7^2 = 147
@@takemyhand1988 no!
open microsoft excel and write that in a cell .
@@LAL666666 7^(1^2) is 7
@@LAL666666 The problem is that you've confused 7^x^2 with 7^2x.
If x=1 then x^2 = 1, not 2. So 7^x^2 = 7^1 = 7. Multiply that by 3 and you get 21.
When you evaluate a "tower" of exponents, you always evaluate from the top downwards.
Start by taking natural logarithms of both sides. ln 7 x^2+ln 3 x- ln 21 =0.
Use quadratic formula we get two roots
x=1. or x=-ln21/ln7 That is all
My kid liked this video. Simple explanation of the laws etc. More videos for kids please.
Can be done by examining the factors of 21 and then solving all the equations? Like the first case with 3^x=3 and 7^x^2=7 and second case with 21 and 1 and then checking the solutions
That approach can yield useful results if you are expecting integer solutions. Unfortunately, in this case one of the two solutions is not an integer.
ln(21)/ln3 = log 21
7
x = 1. Suspiciously easy
I used exactly the same approach ❤❤❤❤❤very nice!!
Very Niice bro❤
or any version of synthetic division; dividing by (x-1); though Vieta's better in this case
Comparing coefficients. 1× log 3 + 1^2 log 7
Yes, could see "1" at just a glance, but knew there had to be a second solution that would be the more difficult, stranger one.
hmm, I got a different approach. I used logarithms as well though. ln[3^x * 7^(x^2)] = ln(21) --> ln(3^x) + ln(7^(x^2)) = ln(21) --> x * ln(3) + x^2 * ln(7) = ln(21) = ln(3 * 7) = ln(3) + ln(7). Put everything on the RHS to the left and group: x * ln(3) - ln(3) + x^2 * ln(7) - ln(7) = 0 --> (x - 1) * ln(3) + (x^2 - 1) * ln(7) = 0 --> (x - 1) * ln(3) + (x - 1)(x + 1) * ln(7) = 0 --> (x - 1)[ln(3) + (x + 1) * ln(7)] = 0.
So either x - 1 = 0 or ln(3) + (x + 1) * ln(7) = 0. So the first one gives x = 1 and the second one gives (x + 1) * ln(7) = -ln(3) = ln(1/3), so x + 1 = ln(1/3) / ln(7) = log_7(1/3) --> x = log_7(1/3) - 1 = log_7(1/3) - log_7(7) = log_7(1/21). This last one is the same as -ln(21)/ln(7), but I just find it a neater answer like this.
on your first line:
7^x^2 is diferent of 7^(x^2)
7^x^2 = (7^x)^2 = 7^(x*2)
@@LAL666666 You say that 7^x^2 = (7^x)^2.
I don't think that's correct. To my knowledge 7^x^2 is indeed 7^x² because if you don't specify the order of exponentiation with brackets you'd start on the right. That is a^b^c^d = a^(b^(c^d)) not a^b^c^d = a^(bcd).
I really like your approach. Despite your name, it's pretty cool and no Quatsch at all! 😛
Personally, I wouldn't turn the logarithmic arguments into fractions or change the bases, at least not until the end, but that's of course a personal preference.
You could then proceed as follows:
x + 1 = -ln(3) / ln(7)
x = -ln(3) / ln(7) - 1 =
-ln(3) / ln(7) - ln(7) / ln(7) =
[-ln(3) - ln(7)] / ln(7) =
-ln(21) / ln(7) = -ln₇(21)
Again, really cool approach!
@@jensraab2902 a^b^c^d = a^(bcd) is correct ! Please, test yourself on Excel.
So SyberMaths,
For many of the final equation results you present, please do complete the arithmetic to show the value of the results.
So here:
-ln21/ln7 = ???
I then need my calculator to find the numerator and the denominator to then show the result, to compare with 1.
Do this for all such results, since you are much more fluid than many of us are.
Thank you
🤓
Cheers
- - -
-ln21/ln7 *is* the value. Any numerical representation would be an approximation.
X= ±1
I've forgotten how to solve this kind of equations. Now, I know how to do it. Nice video sir. Thanks👌🏽
Happy to help
Señores SyberMath. Reciban un cordial saludo, Gracias por este lindo ejercicio, GRACIAS por compartir esa visión que nos permite tener herramientas para afrontar un problema. Felicidades, como la que tengo YO después de ver este desarrollo.
Believe it or not, I solved it exactly the same way.
First, I established the quadratic equation using the ln():
ln(7)*x^2 + ln(3)*x - ln(21) = 0
Then divided by ln(7):
x^2 + ln(3)/ln(7)*x - ln(21)/ln(7) = 0
First, I tried to simplify the discriminant and quadratic formula, but found it too complicatd. Then I remembered that by plugging in, x = 1 is the first solution x1, and I can use Vieta's laws:
x1 * x2 = -ln(21)/ln(7)
So since x1 = 1, we get
x2 = - ln(21)/ln(7) = - log7(21) = -1.564575...
Nice
if x=1 you have 3 * 7^2 = 147
3×7=21
X≈1 . ..
Very interesting.
Glad you think so!
3^x = 21 / 7^x^2
3^x = 3/1^x^2
Because 1^x^2 always 1 in the number and negative number are also going to positive like -1² is 1, -2² is 2 and either, so we can make the solution at below that 1^x^2 is 1 so 3^x = 3
And we can conclude that x = 1
Is it true or not?(Sorry for my bad English)
You can't simplify 21/7^x^2 to 3/1^x^2
What.
x=1
There should be one negative solution apart from your answer 1. 😉😉😉😉😉😉
Bruh
X = 1
I got x=1, which was the obvious.
3^1 x 7^(1x2) = 3 x 49 21
3x7=21 ?
if x=1 then 7^x^2 = 7^2 = 49
7^x^2=7^(x^2)
@@SyberMath No! Please test on Microsoft Excel with real numbers.
Sorry, you can just "see" this one 3^1 * 7^(1^2)= 21
X=zero rejected x =1
x = 1
x=1