วีดีโอ

x^2+x+1=0=>(x-1)(x^2+x+1)=0=>x^3-1=0=>x^6=1ans

x^6=1

X={log7 with base (8/7)}-7.....May be

(x+8)ln7=(x+7)ln8; (ln8-ln7)x=8ln7-7ln8; x=(8ln7-7ln8)/(ln8-ln7)

Solution: 27^x+27^x+27^x = 27 ⟹ 3*27^x = 27 |/3 ⟹ 27^x = 9 ⟹ 3^(3x) = 3² |because of the same base: 3x = 2 |/3 ⟹ x = 2/3

^=read as to the power We know that 1!=1........eqn1 So (1!)^0=1 But It is true that 4!/4=3! 3!/3=2! 5!/5=4! Similarly 1!/1=0! (1!1)^0=(0!)^0........ eqn1 (1)^0=(0))^0 1=(0)^0 Hence 0^0=1...(May be )

2

Enjoyed your content keep growing my fellow Mathematician

Wait - this was a bait and switch - title says 25^x+5^x = 0 and not 20!

This is the most stupid math teacher ever exist...the answer is eiither 0 or 1 or both is trus or both is untrue which in real world now is facing....exist or not exist is in the question.

x^x+x-30=0 x^x+x=30 x^x+x=27+3 x^x+x=3³+3 On comparing x=3

sq root=80^2+3.80+1=6,400+241=6,641 ans .(x=80 and result:x^2+3x+1.).

Nice solution ❤

27ⁿ+27ⁿ+27ⁿ=27 3(27ⁿ)=27 27ⁿ=9 3³ⁿ=3² 3n=2 n=⅔ ❤

0=1

6641

64^x=32^x ; i,e 2^(6x)=2^(5x) ; i,e 6x=5x ; i,e 6x-5x=0; i,e x=0.

Very long-winded method

2^4x = 2^5, 4x = 5, x = 5/4 = 1.25.

x=1.25

Solution is wrong

x=2/3

2/3

2/3

It's uncertain.

First of all: This is no proof. Secondly: Yes, the limit for x->0 for x^x goes to one (which you can prove by writing it as e^(x*lnx) and use l'hopital for the exponent: e^-((1/x)/(1/x²)) = e^x -> 1 But most of all: This only works because you let both go to 0 equally. But you can easily chose a function f that goes to 0 so that f(x)^x does not go to 1. For example for f=e^(-1/x) you get e as limit of f(x)^x. For f=e^(-1/x²) you get infinity. This shows that 0^0 could be different values, depending on how both terms go towards zero. That's why it's not defined.

बहुत महत्वपूर्ण पाठ सर। ❤❤❤❤❤❤❤🎉🎉🎉🎉🎉🎉

Hi

3(3^3x) = 3^3, 3^(3x + 1) = 3^3, 3x + 1 = 3, 3x = 2, x = 2/3.

Correct answer is 4

{{x==2/3},{x==Ln(RealSurd((-3)^2,3))/Ln(3)},{x==Ln((-RealSurd((-1),3)*RealSurd(3^2,3)))/Ln(3)}}

X=2/3

Too easy, I did it in my head,

Xlog8/7=log7/8^7*7=log7/8^7+log7=log7-7log8/7=>x=log7/log8/7-7=1/[log78-1]-7

27^{x+x ➖ }+{27+27 ➖ }=27^{x^2+54}=27^54x^2=1458 100^100^10^40^2^29x^2 10^10^10^10^10^4^10^2^29^1x^2 2^5^2^5^2^5^2^5^2^5^4^2^5^2^1^1x^2 1^1^1^1^1^1^1^1^1^4^1^1^1x^2 2^2^1^1x^2 1^1x^2 1x^2 (x ➖ 2x+1).

you are illiterate and dont how to solve normal roots. your X3 X4 X5 X6 all these four roots are wrong pls first yourself learn to solve and then come hear to teach

3(27^x) = 27, 27^x = 9, 3^3x = 3^2, 3x = 2; x = 2/3

2/3

x=2/3

26.

Stop to send, mathematics . I don't want this came stop it.......

Yes, or ... 7^(x + 8) = 8^(x + 7) 7^x·7^8 = 8^x·8^7 7^8 = (8^x/7^x)·8^7 8^x/7^x = 7^8/8^7 (8/7)^x = 7^8/8^7 x = log(7^8/8^7)/log(8/7) ■ x = 7.572679 🙂

answer is x=4 4^4=256 2^(2×4)=2^8=256

9Y^2 + 8Y -1 = (9Y-1)(Y+1). No need to use the quadratic formula.

Wrong explaination

The correct answer is x=4.

Math olympics for first graders? What is with this yt feed giving me 'olympic' questions that anyone could solge in their head within 3 seconds

You ❤❤❤

Thank you for explaining. From sinx=±1/(√2) , x = ±π/4 (= ± 3.14・・・/4 = ± 0.785・・・ ).

Thank you for explaining. My answer is x = -1, 1. If x=-1, left side is (-1)^(-1) [= (-1)^(odd number) ] = -1 (= right side). Thus, -1 can be one of the solutions.