A Nice Exponential System | Math Olympiads

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Опубликовано:

26 мар 2023

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Комментарии : 41

@seanfraser3125 Год назад

Note that 4^x = 2^x * 2^x, 6^y = 3^y * 2^y, and 10^z = 5^z * 2^z. Using basic exponent rules, the second equation becomes 2^(x+y+z)*(2^x + 3^y + 5^z) = 104 Using the first equation, this is actually 2^(x+y+z) *13 = 104 104/13 = 8, so 2^(x+y+z) = 8. Thus x+y+z = 3

@goldfing5898 Год назад

Divide the second equation by the first equation: 4^x/2^x * 6^y/3^y * 10^z/5^z = 104/13 (4/2)^x * (6/3)^y * (10/5)^z = 8 2^x * 2^y * 2^z = 8 2^(x + y +z) = 2^3 Take the logarithmus dualis: x + y + z = 3

@josukehigashikata1494 Год назад

Divide the two equations to get 2^x . 2^y . 2^z = 8 2^(x + y + z) = 2^3 Therefore: x + y + z = 3

@sscomposer2235 Год назад

I think it is 3. Cos 104/13 is 8 and that becomes 2^(x+y+z)

@shoshosalah3447 Год назад

That's the same I said ❤

@SG49478 Год назад

I like the first method much better. Because the 2nd method assumes that x+y+z is the identical value which we would have to prove. Eventhough the question asks for evaluation of x+y+z we don't know if x+y+z can have different values. In a math olympiad this lack of prove that there is only one possibility for x+y+z would definetly lead to point deduction. The first method includes the prove that 3 is the only possible solution and is therefor the more complete and better one.

@allanmarder456 Год назад

Once get x+y+z=3 you can set solutions in terms of a parameter. Let x=k. From the first equation you have (3^y)*(5^z) = 13/(2^k). y+z=3-k and z=3--k-y So (3^y)*5^(3--k-y) = 13/(2^k). Next lake logs of both sides and solve for y. (its messy and I won't write it here) Then x=k y=solution just derived, and z=3-k-y.

@nasrullahhusnan2289 Год назад

Dividing the second equation by the first one we get 2^x(2^y)(2^z)=8 --> 2^(x+y+z)=2³ --> x+y+z=3

@sonaruo Год назад

dicide both equation and you end up 104/13 equal 2 in the x+y+z take the log of 2 and you got the answer

@kevinmadden1645 Год назад

The Division Property of Equality works quite well here.

@mystychief Год назад

1. xlog2+ylog3+zlog5=log13 2. xlog4+ylog6+zlog10=log104=log4+log13+log2. Put the log13 from 1. en equation 2. Rearranging gives xlog2+ylog2+z(1-log5)=3log2. Everything dividing by log2 gives x+y+z=3.

@leif1075 Год назад

I started this but then finished the way Sean Frasier noted below since second equation 104= 2^×+y+z*13 you have 1o4=13×8 so 8=2^×+y+zand 8=2^3 so we are done..

@usdescartes Год назад

I don't think we were told that x,y,z are Real. If they are Complex, we must take the log of both sides and we x+y+z = 3 + πki/ln√2, where k is an Integer....🙂

@ezzatabdo5027 Год назад

The direct result by following steps is X+Y+Z =3 Why substitute X=0 Thanks sir.

@SyberMath Год назад

Sure. It’s an alternative

@ezzatabdo5027 Год назад

@@SyberMath Thanks.

@mdshamimkhan5731 Год назад

It's easy, put value of z = 0, x = 1 and y=2, in first equation, it will satisfy the condition, so x+y+z = 1+2+0 = 3....😊😊😊

@RichardManns Год назад

The result of the second equation is 8x the result of the first. As you don't have 3 equations, the answer follows.

@wtspman Год назад

Nice!

@SyberMath Год назад

Thanks!

@broytingaravsol Год назад

x+y+z=3

@rakenzarnsworld2 Год назад

2^(2x+y+z)*3^y*5^z=104 3^y*5^z=6.5 x+y+z=3

@yttyw8531 Год назад

Finnaly, I was able to solve equation before syber

@SyberMath Год назад

You beat me!!! Nice 😁🤩💖

@kevinmadden1645 Год назад

Oops! That is how he worked the problem .

@scottleung9587 Год назад

I also got 3 using natural logs.

@SyberMath Год назад

Wow! 😍

@talberger4305 Год назад

@OrenLikes 10 месяцев назад

you should have "copied" 8 down as 2^3 to "save a step"...

@Axman6 Год назад

Do we just ignore that 13 is prime so 2^x+3^y+5^z /= 13? Edit: too much number theory & cryptography, forgot non-integers exist 🙃

@SyberMath Год назад

🙃😊

@shoshosalah3447 Год назад

X+Y+Z=3

@angelamusiemangela Год назад

@alexandermorozov2248 Год назад

Ответ: 3 😃

@BlaqRaq Год назад

I am lazy, so, I don’t like the second method

@SyberMath Год назад

Nice. I'm lazy, too! 😲😜

@derhausfreund5691 Год назад

too easy

@piman9280 Год назад

9+ minutes to explain a problem which can be solved *mentally* by dividing the second equation by the first (in less than 10 seconds) is normally called "milking it."