How to solve an exponential function with quadratic exponent? Reddit r/homeworkhelp
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- Опубликовано: 19 ноя 2024
- How do we solve the exponential equation 16^(x^2+x)=1/2? Remember, if we can make both sides have the same base, then do it! After that, we just have to equate exponents from both sides and solve the resulting question. This question is from Reddit r/homeworkhelp
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For the first time I got the answer by myself!
Congratulations friend
You're my favourite math teacher
Please can you make examples for how to rationalize denominator limits
Oh no my family is held hostage and the only way to help them is to solve this exponential equation
Very informative and easy to follow. Great job at explaining it, helped me a lot
This channel made me solving my math tasks too fast. Now I have nothing to do every school hour when we do revision cause I already solved the revision.
1. 16^(x^2+x)=1/2 -> Given
2. 2^4^(x^2+x)=2^(-1) -> Negative exponent law and expanding
3. 2^(4(x^2+x)=2^(-1) -> Exponent power rule
4. 4(x^2+x)=-1 -> logarithm base 2 on both sides
5. 4x^2+4x+1=0 -> simplify
6. (2x+1)^2=0 -> factor
7. 2x+1=0 -> sqrt 2
8. x=-1/2 -> isolate x
tbh this was a very smooth problem, basic exponential equation + basic quadradic equation and that's it :)
This is such a cool math question I want to show this to my teacher.
Beautiful problem.
Thank You sir
2^(4x^2+4x)=2^-1
same bases
4x^2+4x=-1
x^2+x+1/4
(x+1/2)^2=0
x=-1/2
2^-1 = 1, But 1^-2 = -1.
-1 is not because of 2^-1 but because of some other operation. Find out before simply copying it from other people. Plagiarism
@@sangeetabocki 1 raised to the power of any real number is 1. But here we don't even have any 1^...
@@sangeetabocki me when I spread misinformation:
Amazing video! Man, I've only been noticing this right now...
This was a cute problem.
aw you are correct, this equation is so adorable 🥰
@@sushileaderyt1957You both have a pink profile picture I thought you were the same person.
i also did this but taking ln both sides and make that a quadratic equation .by solving this the avg of its roots is -1/2
-1/2?
What mic are you using sir, name and price
I just guessed numbers. Got it on the third try. -1/2
x²+x-log(16)(1/2)=0
x=-1/2
Can someone tell me what that factoring method is called? What do i search on youtube?
Tic tac toe method.
I like using Po Shen Lo's factoring method. m +/- sqrt(m^2 - p)
m = mean of roots. Equal to -b/(2a) from the quadratic coefficients.
p = product of roots. Equal to c/a from the quadratic coefficients
How he find factor by that method .someone explain
Essentially by guess and check since 4 only has a couple of possible factors. He just didn't show checking 1*4 or negative factors. It's also pretty reasonable that you'd try 2*2 first.
If you want an explanation for the method itself. Basically if a quadratic can be factored it becomes 2 binomials of the form (ax+b)(cx+d), where a and c must multiply to be the quadratic term, b and d must multiply to be the constant term, and the sum of their cross products must be the linear term. So, assuming that your quadratic and constant terms only have a few possible factors each, you can just pick factors for each and check their cross product.
The grid that he makes is just a simple shorthand for checking each combination's product to see if it works. Basically instead of writing your guess as (ax+b)(cx+d) and then doing foil method to check it, you just make a grid under the original quadratic:
a | _ | b
c | _ | d
Then you multiply the diagonals and write them in the middle column and add them to check that the linear term matches. Compared to FOIL method, multiplying the left column is the product of the first terms, multiplying the right column is the last terms, and then multiplying the diagonals are the inside and outside products respectively. We don't need to check the first/last products because we intentionally choose factors or the original terms, it's just the sum of the diagonals that we need to check.
If our choice works, then we have a valid factorization. If it doesn't, then we try another. If we run out of factor combinations, or there are too many to check them all, we need to move on to another method of solving it, such as quadratic equation.
Here you know (ax + 1)(bx + 1) = abx² + (a + b)x + 1 = 4x² + 4x + 1.
So, ab = 4 and a + b = 4.
So, can you guess the answer now?
If not, try a = 1 and b = 4. This fails. Then they a = 2 and b = 2 and this works.
So (2x + 1)(2x + 1) = (2x + 1)² =
4x² + 4x + 1
Solution:
1/2 = 2⁻¹
16 = 2⁴
so the equation becomes
(2⁴)^(x²+x) = 2⁻¹
2^(4x² + 4x) = 2⁻¹ |log₂
4x² + 4x = -1 |:4
x² + x = -1/4 |+1/4
x² + x + 1/4 = 0
x² + 2 * 1/2 * x + (1/2)² = 0
(x + 1/2)² = 0 |√
|x + 1/2| = 0
The only value that satisfies this equation is x = -1/2
To start with, 2 raised to the power of -1 is 1.
1/2 =2^1 is correct in this case because 1/2 is 1. How are you getting it in - ?
1 raised to the power of -2 will be -1.
Also squaring 1/2 we get 1/4 or simply putting it squaring .5 we will get .25.
@@sangeetabocki I don't even know to begin with this nonsense. Except for the last sentence, everything you said is utterly wrong. And that sentence is irrelevant, because I actually use 1/4 = (1/2)², so the reverse of what you stated.
@@sangeetabocki Just to give you the 'power rule' required here: (This a given LAW/RULE, that is immutable!)
a⁻ⁿ = 1/(aⁿ)
as such:
1 ≥ a⁻ⁿ > 0
(can NEVER be negative!)
And I have absolute no idea, where you get that 1/2 = 1. That is literally a contradiction!
@@m.h.6470 1. What is 1^-2 ? For the contradiction refer to the one who devised this equation. There is some technical error which I am trying to reiterate here.
@@m.h.6470 The rest is also not non sense it is maths. Objective not subjective. Check for yourself.
4x² + 4x = -1
Later down the line
2x = -1 😂
To check equation, If we assume x = 1 then 16^1^2+1 = 1/2
16^1+1 = 1/2
16^2 =1/2
256 = 1/2
Checking, 256 x 2 = 1
If, 512 = 1
Then 256 x 2 = 512
Or 256 = 512/2
Dividing both by 1000 .25 = .51/2
Or .25x2 = .5
.5 = .5
Also, 16^2=1/2
256= 1/2
256 x 2 = 512
Using approximations, 250 = 500/2 or 500 = 1000/2.
Computing the value of x in the equation the answer comes around .0046 so rounding off the same we can say it is .5 which is equivalent to 1/2. I was wondering if this equation could be reduced to 4^x+√x= 1/2.
❤
16^(x²+x) = 1/2
2^4(x²+x) = 2^(-1)
By identification
4(x²+x) = -1
4x² + 4x + 1 = 0
4(x² + x + 1/4) = 0
(x + 1/2)² = 0
x + 1/2 = 0
x = -1/2
Easy grade 9-10 level
Technically, 2^-1 = 1. Not in negative. Also, you need to solve the equation with your solution to check. See if it fits.
@@sangeetabocki Did you pass middle school? a^-n = 1/a^n
@@UserSams-ve2mj if a^- n = 1/a^n, then In this situation, if a= 2 and n = 1. Then, 2^-1 = 1/2^1 which results in 1 = 1/2. You just validated my point. Thanks. By the way, I managed to pass middle school but I was caught cheating (Farah)in divinity exam. If that is a red flag, then no wonder I feel divine so less.
@@sangeetabocki And from where did you even get the 1 = 1/2? Go focus on your math class kid
@@UserSams-ve2mj my car number should be 5025 as 1=1/2. 50=25
16^(x²+x) = 1/2
(2^4)^(x²+x) = 2^(-1)
2^(4x²+4x) = 2^(-1)
4x² + 4x = -1
4x² + 4x + 1 = 0
x² + x + 1/4 = 0
x² + 2x/2 + 1/4 = 0
(x+1/2)² = 0
x = -1/2
16^((-1/2)²+(-1/2)) = 1/2
16^(1/4-1/2) = 1/2
(2^4)^(-1/4) = 2^(-1)
2^(-1) = 2^(-1) ✓
Or
16^(x^2 +x) = 1/2
1/2 = 16^-1/4
-1/4= X^2 + X
X^2 + X + 1/4 = 0
x= - 1/2
Pick up your scientific calculators and compute 2^-1 = 1
Listen carefully to the question either make both the sides 1/2 or find the value of x. 2^-1 = 1 not 1/2.
In the real life application the equation wants to convey that left side is equal to half of right. In that case, applying with the value of 1 we may say that -1/2 is equal to 1/2 half - half = zero or half + half = one balancing both sides.
1/4 is equal to half of 1/2. .25 is equal to .5.
Also, 16^2=1/2
256= 1/2
256x2 = 1
512 = 1
Balancing both side, 256 is approximately half of 500 and 512 is approximately half of 1000.
Try solving this equation with x = 1/4 or .25 you will get better near approximation.
100th comment
Grade 12? I was doing this in the 3rd grade! Wow.
Algebra and exponents in 3rd grade? Did you go to private school
Now you're just lying
Not that early. We learnt exponents in 7th grade and quadratic equations in 9th grade( I'm in 9th currently).
@@BhumanBudhirajaI feel 9th is a little late, but it really depends on the country and region. I’m from the us and we did quadratics in 7th grade
X=-1/2