16

1,5

1/3

log(2)5

16

I get two more as the solns to x = +/- e^(ln(2)/4x) . The negative version has one soln, and the positive has another 2 more, of those 2, one is 16. x* = { -4 W(-ln(2)/4)/(ln(2)), -4 W(ln(2)/4)/(ln(2)), 16}

Why the solution is so complicated? X^3+X=130 X^3+X^1=5^3+5^1 /*did you find it by a chance?*/ X=5 2^n=5 n=Lg2(5)

@matholympiadonlinesolver 1/ 2 log base 5 = 5 log base 2 (= 2,321928) applied in the initial equation does not prove to be correct. x = 3 and - 0,430676 (= negative 0,430676) are correct solutions and they give the result 500.

2005 is the answer,I solved it now

3/2

1/3

1/3

7! is the answer,solved in mind

You only found one solution, there are other real solutions

Verme do capeta

Gud👍👍

And the other 3 Solutions???

X =3 y=1

5^x+5^(x+2)=52 5^x+25(5^x)=52 26(5^x)=52 5^x=2 x=log_5(2) ❤

3^(x^2) = 81^(x - 1) 3^(x^2) = (±3)^[4(x - 1)] 3^(x^2) = 3^[4(x - 1)] or 3^(x^2) = (-3)^[4(x - 1)] x^2 = 4x - 4 3^(x^2) = (3e^(i.π))^[4(x - 1)] x^2 - 4x + 4 = 0 3^(x^2) = 3^[(1+i.π/ln(3))4(x - 1)] (x - 2)^2 = 0 x^2 = 4(1 + i.π/ln(3))(x - 1) x = 2 usw. usw. _________ usw.

52

27^x = x^-1 Ln x / x = -ln 27 x^-1 ln x^-1= ln 27 e^lnx^-1* lnx^-1=ln27 W(e^lnx^-1* lnx^-1)=W(ln27) Ln x^-1= W(ln27) x^-1= e^w(ln27)= 3 x = 1/3

x Ln27= W(3Ln3)= Ln3 x=⅓

X={3 ; -0.43}

xln5 + (1/x)(x - 1)ln8 = ln500 x²ln5 + (x - 1)ln8 - xln500 = 0 x²ln5 + xln(8/500) - ln8 = 0 x²ln5 - xln(500/8) - ln8 = 0 x²ln5 - xln(125/2) - 3ln2 = 0 x = [ln(125/2) ± √(ln²(125/2) + 12ln5ln2)]/(2ln5) ln²(125/2) + 12ln5ln2 = ln²125 + ln²2 - 2ln125ln2 + 12ln5ln2 = ln²125 + ln²2 - 2ln125ln2 + 4ln125ln2 = (ln125 + ln2)² x = [ln(125/2) ± (ln125 + ln2)]/(2ln5) x = 2ln125/(2ln5) => *x = 3* x = -2ln2/(2ln5) => *x = -ln2/ln5*

I will make vedio on the same problem thanks for the inspiration

Hello, I have noticed a problem. The thumbnail of the video shows a minus between the exponential numbers, but in the video you solve it with a plus.

Unnessory leanthy

I like this solution. make more videos

Cool

m = 2004 and n = 2005 are obvious solutions.

4009 is the other trivial solution

What age Olympiad would that one be in? 60 yr old woman who did maths A level at school 40 years ago, feeling quite pleased to get this one easily.

With small numbers like this, you can just list perfect squares under 45 and find by try and error. I solved it in 15 seconds like this :))

Might I ask what country are you from?

Tried calculating based off the thumbnail, i got c²-a² = -5 so (c+a)(c-a) = -5, using that and a little brute force, the 4 possible values are c = ±2, a = ±3. From there you get b = ±6 which is the solution

Always a+b > a-b , so you should not have four cases . I will assign the greater factor to the greater eqn

writing z for sin^2 ( x) one gets f ( x) = (1 - z + z^2) /( z + ( 1 - z) ^2) = (1 - z + z^2) /( 1 - z + z^2) = 1

How do you pronounce x?

One other way to see it is to know that sin^4 -cos^4= sin^2-cos^2. Then everything cancels out and you get the result.

x is 3

I did f(x) = (cos²x + sin²x.sin²x)/(sin²x + cos²x.cos²x) = (cos²x + sin²x.(1 - cos²x))/(sin²x + cos²x.(1 - sin²x)) = (cos²x + sin²x - sin²xcos²x))/(sin²x + cos²x - cos²xsin²x)) = (1 - sin²xcos²x)/(1 - sin²xcos²x) = 1. Very easy.

98

24

You messed up the x part, you somehow made it into /2(sqrt(5)+2) instead of 2(sqrt(5)+4)

derivative of f(x) is 0. f(x) is constant and f(0)=1. hence f (x)=1 for all x.

Bro.. Nice explanation

So simple Well done

first like

But arent there infinite solutions such as logbase2(10) for x and logbase3(3) for y or whatever other numbers you want that make 7