This EQUATION is as X^p=1 So X^p=1 in R • base X=1 and exponent p in R, • or X=-1 and p even integer , • or p=0 and X0: See origine here: th-cam.com/video/aGBe9ObqlwA/w-d-xo.html
How to Solve: When we have an exponent equal to 1, for example a^b=1, then there are only two possibilities: the base is 1 or the power is 0, so when a^b=1, a=1 or b=0! We also need to plug in our possible values of x, in case there are any extraneous solutions. Using this knowledge, the problem is quite straightforward! Our possibilities are 2x+6=1 and x^2-9=0. Solve both equations for x: 2x+6=1 -> 2x=-5 -> x=-5/2=-2.5=-2 1/2. x^2-9=0 -> x^2=9 -> x=+/-3 So our possiblities are x=-5/2 and x=+/-3. All are valid except x=-3, which, when substituted in, produces 0^0, an indeterminate expression. Therefore, the correct solutions for the value of x is x=3 and x=-5/2.
One other possibility that needs to be checked is for the base to be equal to -1 and the exponent is also an even integer. For the base to be -1, x would have to be -7/2. But if x = -7/2, the exponent would not be an even integer. So there is no additional solution from this path, but it's an option that needed to be checked.
This EQUATION is as X^p=1 where X=2x+6 and p=x^2-9 So X^p=1 in R • base X=1 and exponent p in R: 2x+6=1 and x^2-9 in R ===>x=-5/2 is solution • or X=-1 and p even integer: 2x+6=-1 and x^2-9 even: no solution • or p=0 and X0: x^2-9=0 and 2x+60 ==> x=3 solution but x=-3 is not solution bcse 2x+6=0. See origine here: th-cam.com/video/aGBe9ObqlwA/w-d-xo.html
If the base was 3x-12 then it will be -1 when x=11/3. The exponent will be 40/3 So (-1)^(40/3) = ((-1)^1/3)^40 =(1-)^40 =1. Even if you take 40th power first and then cubic root the same result. Basically when the base equals to -1 the exponent has to be a rational number with even numerator which include even integers.
take both sides as log_10 and (x^2 - 9 ) log_10 (2x+6)=log_10 ( 1 ) =>> (x^2-9)log_10 (2x+6) =0. notice should 2x+6 > 0 or x > -3. next first step x^2-9 =0 === > x=3 or x+-3 so x=-3 nor sulution. second step log_10 (2x+6) = 0 means 10^0=2x+6 ====> 2x=-5 and x = - 2.5 which in domain of x>-3 resukt is x=3 or x=-2.5
For 3-rd case (b=-1) a is also can be rational non-integer, if a nominator is even and a denominator is odd. But as I know, the negative base for non-integer exponent is'n recommended...
I think you are correct. If the base was 3x-12 then it will be -1 when x=11/3. The exponent will be 40/3 So (-1)^(40/3) = ((-1)^1/3)^40 =(1-)^40 =1. Even if you take 40th power first and then cubic root the same result. Basically when the base equals to -1 the exponent has to be a rational number with even numerator.
By inspection, either the exponent is 0 or the base is 1, or -1 if the exponent is even. So +/-3 or -5/2 but not -7/2. In fact not -3 either because the base will be zero. Haven't looked at complex solutions.
From the positive side it's pretty straightforward, because then the base of the exponent is positive and you just use L'Hôpital's rule. The negative side is trickier, because then you're dealing with complex numbers. You can still use L'Hôpital's rule, but you gotta be careful with the branch of the power function you're taking.
Think of tangent 90 degrees. y= tan x as a function, you will notice that an asymptote occurs at 90 degrees and as x approaches the asymptote the values for y are totally different for each side. Well, a similar situation exists in this case. The graph plot shown in the end here demonstrates that.
0^0 is undefined. There is no continuous extension of x^y as x, y both go to zero. If x=0 and y varies, the function is 0. If y=0 and x varies, the function is 1.
Pourquoi il masque certains commentares ? Sûrement pour masquer son copiage! Comme ici: X^p=1 in R • base X=1 and exponent p in R, • or X=-1 and p even integer, • or p=0 and X0, See origine here: th-cam.com/video/aGBe9ObqlwA/w-d-xo.html
You get plenty of page views without falsely representing this problem as Olympiad level in difficulty. B^0 = 1 and 1^n = 1 is all that is required. Still don't get how you get so much attention by working the same type of of small "nice" constant type problems. You literally have 7-8 problem types that you rework ad infinitum. I guess no need to expand your math knowledge. It must be your smooth psychiatrist couch voice.
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Bunch of cases to consider. Case I: Exponent = 0. Case II: 2x+6 = 1 and exponent = 1. Case III: 2x+6 = -1 and the exponent is an even natural number.
This EQUATION is as X^p=1
So X^p=1 in R
• base X=1 and exponent p in R,
• or X=-1 and p even integer ,
• or p=0 and X0:
See origine here:
th-cam.com/video/aGBe9ObqlwA/w-d-xo.html
"Bunch" must mean something different to you.
Easy mode:
log both side ---> (x²-9)ln(2x+6) = 0 (any log of 1 = 0)
expend ---> (x+3)(x-3)ln(2x+6) = 0
check for the zero solutions :
---> ln(2x+6) = 0 so 2x+6 = 1 ---> x = -5/2
---> (x-3) = 0 ---> x = 3
---> (x+3) = 0 ---> plug it; gives 0⁰ ---> not a solution
Done!
With log some solutions are lost as in:
(2x-18)^(x^2-9)=1 ===>
(x^2-9)ln(2x-18)=0 ===>for x=-3 and x=3, ln(2x-18) is not defined bcse 2x-18
logarithms completely unnecessary. Just check the cases B^0 = 1 and 1^n = 1. (Easier mode)
How to Solve: When we have an exponent equal to 1, for example a^b=1, then there are only two possibilities: the base is 1 or the power is 0, so when a^b=1, a=1 or b=0! We also need to plug in our possible values of x, in case there are any extraneous solutions. Using this knowledge, the problem is quite straightforward!
Our possibilities are 2x+6=1 and x^2-9=0. Solve both equations for x:
2x+6=1 -> 2x=-5 -> x=-5/2=-2.5=-2 1/2.
x^2-9=0 -> x^2=9 -> x=+/-3
So our possiblities are x=-5/2 and x=+/-3. All are valid except x=-3, which, when substituted in, produces 0^0, an indeterminate expression.
Therefore, the correct solutions for the value of x is x=3 and x=-5/2.
One other possibility that needs to be checked is for the base to be equal to -1 and the exponent is also an even integer. For the base to be -1, x would have to be -7/2. But if x = -7/2, the exponent would not be an even integer. So there is no additional solution from this path, but it's an option that needed to be checked.
This EQUATION is as X^p=1 where X=2x+6 and p=x^2-9
So X^p=1 in R
• base X=1 and exponent p in R:
2x+6=1 and x^2-9 in R ===>x=-5/2 is solution
• or X=-1 and p even integer:
2x+6=-1 and x^2-9 even: no solution
• or p=0 and X0:
x^2-9=0 and 2x+60 ==> x=3 solution but x=-3 is not solution bcse 2x+6=0.
See origine here:
th-cam.com/video/aGBe9ObqlwA/w-d-xo.html
@@mdyankee55 Smart. Thanks for reminding me that.
×=3, x =-5/2.
If the base was 3x-12 then it will be -1 when x=11/3. The exponent will be 40/3
So (-1)^(40/3) = ((-1)^1/3)^40
=(1-)^40 =1. Even if you take 40th power first and then cubic root the same result.
Basically when the base equals to -1 the exponent has to be a rational number with even numerator which include even integers.
take both sides as log_10 and (x^2 - 9 ) log_10 (2x+6)=log_10 ( 1 ) =>> (x^2-9)log_10 (2x+6) =0. notice should 2x+6 > 0 or x > -3. next first step x^2-9 =0 === > x=3 or x+-3 so x=-3 nor sulution. second step log_10 (2x+6) = 0 means 10^0=2x+6 ====> 2x=-5 and x = - 2.5 which in domain of x>-3 resukt is x=3 or x=-2.5
Have you ever tried to solve this equation? It is a tricky one...
x^((x-1)^2) = 2*x + 1
For 3-rd case (b=-1) a is also can be rational non-integer, if a nominator is even and a denominator is odd. But as I know, the negative base for non-integer exponent is'n recommended...
I think you are correct. If the base was 3x-12 then it will be -1 when x=11/3. The exponent will be 40/3
So (-1)^(40/3) = ((-1)^1/3)^40
=(1-)^40 =1. Even if you take 40th power first and then cubic root the same result.
Basically when the base equals to -1 the exponent has to be a rational number with even numerator.
If you’re using an iPad, that extra line at the end is a glitch that happens if you are charging the iPad so you may want to keep that in mind.
For the 3rd case, what about rational numbers with even numerator and odd denominator for the exponent?
*Yeah, but 49 is clearly odd and 4 is clearly even.*
For 2x+6>=0, take log and find some answer.
For 2x+6
By inspection, either the exponent is 0 or the base is 1, or -1 if the exponent is even. So +/-3 or -5/2 but not -7/2. In fact not -3 either because the base will be zero. Haven't looked at complex solutions.
x=3, -5/2
X²-9 =0
X= 3 or x=-3
or you can just multiply both sides with 0 and always get the correct answer😊
X=3
Solved by inspection. X= 3,-3,-2.5
-3 results in 0^0
@@renfreak Didn't see that initially. Thanks
Why are some comments are hidden ?
x = *3* or -3 or *-5/2*
X= -2.5, 3
x=3
well done :)
What do you use to do your presentation (what software)?
What is the limit as x approaches -3? Does it matter if it approaches from the negative or positive side?
From the positive side it's pretty straightforward, because then the base of the exponent is positive and you just use L'Hôpital's rule.
The negative side is trickier, because then you're dealing with complex numbers. You can still use L'Hôpital's rule, but you gotta be careful with the branch of the power function you're taking.
Think of tangent 90 degrees. y= tan x as a function, you will notice that an asymptote occurs at 90 degrees and as x approaches the asymptote the values for y are totally different for each side. Well, a similar situation exists in this case. The graph plot shown in the end here demonstrates that.
Why is the answer x = -3 not acceptable? After all, 0 ^0 = 1.
0^0 is undefined. There is no continuous extension of x^y as x, y both go to zero. If x=0 and y varies, the function is 0. If y=0 and x varies, the function is 1.
@@rickdesper Thank you!
But what about the well-known expressions: (0^O)^0 = 1^0 = 1, and: 0^(0^0) = 0^1 = 0?
@levskomorovsky1762 they're wrong
x=plus ou moins 3
Pourquoi il masque certains commentares ?
Sûrement pour masquer son copiage!
Comme ici:
X^p=1 in R
• base X=1 and exponent p in R,
• or X=-1 and p even integer,
• or p=0 and X0,
See origine here:
th-cam.com/video/aGBe9ObqlwA/w-d-xo.html
میتونین هم از همون اول کاراز دو طرف لگاریتم بگیرید تا جواب ها بدست بیاد
Why are some comments hidden ?
X^p=1 in R
• base X=1 and exponent p in R,
• or X=-1 and p even integer,
• or p=0 and X0,
See origine here:
th-cam.com/video/aGBe9ObqlwA/w-d-xo.html
@@Frank-kx4hcim from iran
@@Frank-kx4hc
If you take the logarithm of both sides of the equation, the answer will be much easier
log 1 =log(2x+6)^x^2-9
I am there's
You get plenty of page views without falsely representing this problem as Olympiad level in difficulty. B^0 = 1 and 1^n = 1 is all that is required.
Still don't get how you get so much attention by working the same type of of small "nice" constant type problems. You literally have 7-8 problem types that you rework ad infinitum. I guess no need to expand your math knowledge.
It must be your smooth psychiatrist couch voice.