@@MrMartinSchou each squaring removes one root and frees one x. But one has to be careful that those have to be included in the next squaring. 1. x*tripple SQRT 2. x^3 * double root 3. x^7 * single root 4. x^15 on the right side it's 5^2^2^2^2=5^16
Expand from the inside out: x * sqrt(x) = x^(3/2) sqrt(x^(3/2))) = x^(3/4) x * x^(3/4) = x^(7/4)... Rinse and repeat until x^(15/16) = 5, so x = 5^(16/15)
We'd write the result as 5× (15)√5 (five times 15th root of five), the last reformulation being x^15 = 5^16 --> x^15 = 5^15 × 5 --> 15th root both sides, x = 5 × 5^(1/15) or 5× 15√5
More interesting question: solve for x>0 in *√(x√(x√(x√(x√(x√(x√(x....)))))))=5* , _where the dots extend indefinitely._ Solution: the function f(x) on the left hand side satisfies the functional equation √(xf(x))=f(x). Squaring and solving for f(x) gives that either f(x)=0 identically or f(x)=x identically. The first solution is incompatible with the equation. The second solution gives x=5.
@Coy33-co2tq I’m guessing you don’t have a maths background or are young. The fundamental theorem of algebra states that there are n complex solutions for an nth order polynomial. You only have one of the 15 solutions
Both methods use different properties of exponents . It's not just about the representation of the answer is same . So both the method are practically different.......
@@ruchakhokale1852 generally when we talk about different methods to solve a question, we mean entirely different approach, i.e. using different concepts. Like some questions can be solved with quadratic formula, or spliting middle term, cooridinate geo and etc. While in this case, its practically the same way. As i said, just different way of writing
Square it 4 times and get x^15 = 5^16 => x = 5^(16/15)
And what, exactly, did you square 4 times to end up at 15?
@@MrMartinSchou each squaring removes one root and frees one x.
But one has to be careful that those have to be included in the next squaring.
1. x*tripple SQRT
2. x^3 * double root
3. x^7 * single root
4. x^15
on the right side it's 5^2^2^2^2=5^16
Expand from the inside out:
x * sqrt(x) = x^(3/2)
sqrt(x^(3/2))) = x^(3/4)
x * x^(3/4) = x^(7/4)...
Rinse and repeat until
x^(15/16) = 5, so
x = 5^(16/15)
This question was so easy
We'd write the result as 5× (15)√5 (five times 15th root of five), the last reformulation being x^15 = 5^16 --> x^15 = 5^15 × 5 --> 15th root both sides, x = 5 × 5^(1/15) or 5× 15√5
How do we justify the de-nesting in second method?
(x ➖ 5x+5).
Me adding and multiplying exponents at the left: 🤡
Even tho it is ok... It is more complicated xD
√(x√(x√(x√x))) = 5
x√(x√(x√x)) = 5² --- ^2
x²(x√(x√x)) = 5⁴ --- ^2
x³√(x√x) = 5⁴
x⁶(x√x) = 5⁸ --- ^2
x⁷√x = 5⁸
x¹⁴(x) = 5¹⁶ --- ^2
x¹⁵ = 5¹⁶ = 5(5¹⁵) --- ¹⁵√
x = 5(¹⁵√5)
when I first saw the question my immediate reaction was that xsqrt(x) = 3/2, thus I used the second method, very nice video though
√(x√(x√(x√x))) = 5
x√(x√(x√x)) = 5^2
(x^3)(√(x√x)) = 5^4
(x^7)(√x)) = 5^8
x^(15) = 5^(16)
x = 5^(16/15)
More interesting question: solve for x>0 in *√(x√(x√(x√(x√(x√(x√(x....)))))))=5* , _where the dots extend indefinitely._
Solution: the function f(x) on the left hand side satisfies the functional equation √(xf(x))=f(x). Squaring and solving for f(x) gives that either f(x)=0 identically or f(x)=x identically. The first solution is incompatible with the equation. The second solution gives x=5.
There are 15 solution
No, there are infinite solutions, because we can add 1 or 2 or 3... until infinite and by both sides etc...
@Coy33-co2tq??? No? There are 15 solutions because it’s a 15th order polynomial
@admiraloctavio5860 I have solution 5^16/15
@Coy33-co2tq I’m guessing you don’t have a maths background or are young. The fundamental theorem of algebra states that there are n complex solutions for an nth order polynomial. You only have one of the 15 solutions
@@admiraloctavio5860
Are you saying there are 15 ways to solve the math or are you saying there are 15 answers
I am not trying to be rude but both the methods are practicing same, just different way of representing lpl
Both methods use different properties of exponents . It's not just about the representation of the answer is same . So both the method are practically different.......
@@ruchakhokale1852 generally when we talk about different methods to solve a question, we mean entirely different approach, i.e. using different concepts.
Like some questions can be solved with quadratic formula, or spliting middle term, cooridinate geo and etc.
While in this case, its practically the same way. As i said, just different way of writing
(x^15)^(1/16)=5
[(x^15)^(1/16)=5]^16
x^15=5^16 , x=5(5)^(1/15)
Please talk a little faster so I have no chance of understanding you.
Thank you.
X=5¹⁶
forgot it's x^15 and not x on the LHS
You american guys call it higher mathematics? 😂😂
the channel's name is "higher mathematics", and I'm pretty sure that every american guy didn't create this channel
First method was a lot better and the easiest way to go through things.
But I learned something from the second one about roots!!