Working with an Exponential System

[@christianthomas9863]

just multiply the two expressions member to member: you get 2^(2x) = 5^5 ; next do divide the exressions member to member and you get: 2^(x+y) * 2^(y-x) = 5 after logging thoses two équations and divide member to member you got: x/y = 5, so 2^(x/y)=32

[@mcwulf25]

Pretty much what I did

[@lucafumagalli1829]

I did the same, much faster...

[@yanceyward3689]

Exactly the way I did it, too.

[@christianthomas9863]

@@mcwulf25

[@RAG981]

Two minutes. I totally agree. Seemed the obvious way before he said anything. Well done.

[@devondevon4366]

2^x + y = 125 given 2^x+y =5^3 2^x-y = 25 given 2^x-y =5^2 2^x+y+(x-y) =5^5 multiply both equation 2^2x = 5^5 equation A 2^x+y-(x-y =5 divide both equation 2^2y = 5 equation B 2^2x/5 = 5 raised equation B to the power of (1/5) equation C 2^2x/5 = 2^y since equation C and B = 5 2x/5 =2y, equating the base 2x =10y x = 5y hence 2^x/y = 2^5y/y = 2^5 =32 answer

[@TheShacharZiv]

Since 125= 5^3 and 25=5^2, I raised the first equation to the second power and the second equation to the third power. From here: 2(x+y)=3(x-y), from here x=5y, 2^5=32

[@misterdubity3073]

I did 2nd method in my head. Nice problem. Congratulations on 100K followers!

[@SyberMath]

Awesome! Thank you! 🥰🤗💖

[@佐藤広-c4p]

In short, since it is enough to calculate x/y from x+y and x-y, I think that the 1st method of finding x and y respectively using logarithm and then calculating x/y is the easiest and easiest to understand.

[@DrQuatsch]

I sort of combined both methods. I first multiplied eq. 1 and eq. 2 to get 2^2x = 5^5. And then I divided eq. 1 by eq. 2 to get 2^2y = 5. From 2^2x = 5^5 it follows that 5 = 2^(2x/5). Now we have both that and 2^2y equal to 5, so they have to be equal to each other. So 2^2y = 2^(2x/5). Same base, so exponents need to be equal; 2y = 2x/5. Multiply both sides by 5/2y to get 5 = x/y. So 2^(x/y) = 2^5 = 32.

[@ranjithraom]

Dividing 2^(x+y) by 2^(x-y) one will get the solution 2^(2y)=5. In the equation 2^(x-y)=25=5^(2). By Replacing 5 by 2^(2y) the final equation will be x-y=4y which means x=5y or x/y=5. So 2^(x/y) = 2^(5) = 32.

[@andy_in_colorado7060]

Is the answer 32? Going to watch now and see if I did it in my head correctly.... Yay it is! I used base-5 logs instead of straightforward base-2, because I wanted to turn 125 & 25 into 3 & 2 instead of multiplying in my head, but it made it more convoluted than the first method. (Not to mention, I didn't really need to worry about the base of the log to extract the 3 & 2).

[@imonkalyanbarua]

I did it by the first method but i am very impressed by the simplicity and the beauty of the second method. Thank you. 😇🙏

[@SyberMath]

Most welcome 😊

[@imonkalyanbarua]

@@SyberMath 😇🙏

[@Mekenyejustus]

x+y=(2ln5)/ln2 and x-y=(2ln5)/ln2 so that 2x=5ln5/ln2 and 2y=ln5/ln2 Hence x/y=2x/2y=5ln5/ln2.ln2/ln5=5 2^(x/y)=2^5=32

[@vishalmishra3046]

x + y = log2(125) = 3 log2(5) and x - y = log2(25) = 2 log2(5) On adding, 2 x = (3 + 2) log2(5) and on subtracting 2 y = (3 - 2) log2(5) On dividing, x / y = 2x / 2y = 5 / 1 = 5 Therefore, 2^(x/y) = 2 ^ 5 = 32. *Simple* right ?

[@bryantwiltrout5492]

I used the 2nd method. It was pretty easy to figure out really.

[@SyberMath]

Nice job!

[@seegeeaye]

From given equations we have 2^(2x)=5^5 and 2^(2y)= 5, so x=5y, so 2^(x/y)=32

[@MrLidless]

Much quicker to multiply the two equations to get 2²ˣ = 5⁵, divide them to get 2²ʸ = 5, log both, and immediately get x/y = 5.

[@loohooi6545]

Yes,I agree,but actually you no need to log both,just substitite 5 with 2^2y into the first equation is a better way.

[@MrLidless]

@@loohooi6545 Both quick and equivalent routes to the same destination.

[@SuperYoonHo]

Thank you so much SyberMath sir!!! YOur getting more subscribers I hope!

[@SyberMath]

Thank you too! 🥰🧡

[@SuperYoonHo]

@@SyberMath It's ok sir!👌👌🙏🙏

[@gerhardb1227]

EQ1: 2^x * 2^y = 125 EQ2: 2^x/2^y = 25 EQ1: 2^x = 125/2^y put EQ1 in EQ2 125/2^y/2^y = 25 2^y^2 = 5 => 2^y = 5 put result in EQ1 2^x * 5 = 125 => 2^x = 25 2^x/2^y = 2^(x/y) = 2^(25/5) = 2^5 = 32

[@maxm9960]

you can take natural log on both sides but not solve to the end. (x+y)ln2 = 3ln5 -------- (1) (x-y)ln2 = 2ln5 ----------(2) (1)-(2) --> 2yln2 = ln5 -> yln2 = (1/2)ln5 ------------ (3) then (2)+3) -->xln2 = 2ln5 + (1/2)ln5 = (5/2)ln5 -------------(4) then (4)/(3), LHS = x/y, RHS = (5/2)ln5 / (1/2)ln5 = 5, i.e. x/y=5 hence 2^(x/y)=32

[@SyberMath]

Nice!

[@fernandoderoque1920]

Considering these 2 equations Multiplying 2^(2x)=5^5 Dividing 2^(2y)=5 Logging these 2 new equations base 5 results 2x log5(2)=5 2y log5(2)=1 Dividing both results x/y = 5 Then 2^(x/y) = 32

[@Packerfan130]

2^(x + y) = 125 2^(x - y) = 25 2^(2x) = 2^(x + y) * 2^(x - y) = 125 * 25 = 5^3 * 5^2 = 5^5 2^(2y) = 2^(x + y) / 2^(x - y) = 125 / 25 = 5^3 / 5^2 = 5 2x * ln 2 = ln 5^5 = 5 * ln 5 2y * ln 2 = ln 5 x/y = (2x ln 2) / (2y ln 2) = (5 ln 5) / ln 5 = 5 2^(x/y) = 2^5 = 32

[@cb_q]

cool... 2nd method really cool!

[@mcwulf25]

My method more like #2 but a bit simpler. Multiply together and 2^2x = 5^5 Divide and we get 2^2y = 5, or 2^y = 5^(1/2) If u = 2^(x/y) then u^y = 2^x = 5^(5/2) = (2^y)^5 = (2^5)^y Then take the yth root to get 2^(x/y) = 2^5 = 32

[@ManjulaMathew-wb3zn]

Next method Multiply equations 2^2x=.5^5 Divide them 2^2y=5 Substituting in first equation for 5 in terms of y 2^2x = (2^2y)^5 Taking 2y th root on both sides 2^(x/y)=2^5=32

[@Desam1000]

There is another (in my opinion shorter) way by first solving to x=5*log4(5) and y=log4(5)

[@nicogehren6566]

good question

[@SyberMath]

Thanks!

[@yanceyward3689]

I really liked the 2nd route.

[@halitiskender1324]

Anlatımın ve sesin çok güzel

[@SyberMath]

Tesekkur ederim!

[@MallorcaDuck]

chris cross apple sauce XD Now that I laughed at it I feel like I'm fluent in english

[@giuseppemalaguti435]

Lavorando con i logaritmi ottengo x/y=5,percio ottengo 2^5=32

[@anupchattaraj5039]

The 2nd method is very nice.

[@SyberMath]

Glad to hear that!

[@SimonClarkstone]

I did: Define k such that 5^k = 2. then 5^(kx+ky) = 5^3 and 5^(kx-ky) = 5^2 so kx+ky = 3 and kx-ky =2 obviously kx = 2.5 and ky = 0.5 x÷y = kx÷ky = 5 2^5 = 32

[@SyberMath]

That's very good! You're kind of logging both sides with base 5

[@xarax7950]

EASY !!!

[@chiragsharma8395]

Attending premier for the first time

[@morteza3268]

This is how I solved it; 2^(x+y)=5³ and 2^(x-y)=5² Multiply two equations once and divide once. Eq1)2^(x+y)×2^(x-y)=5⁵=2^(2x) Eq)2^(x+y)/2^(x-y)=5=2^(2y) so 5^(1/y)=4 We know ; (4)^x=5^5 Replacement 4=5^(1/y) So 5^(x/y)=5^5 x/y=5 2^(x/y)=32

[@morteza3268]

2^(x+y)=5³ 2^(x+y-2y)=5²=5³/2^(2y) 5^(1/y)=4 2^((x+y-y)/y=(5^(3/y))/2=2^5

[@mamadoudiop1958]

Merci beaucoup

[@SyberMath]

Je vous en prie!

[@БашШа]

Легко и просто решает наш заграничный блогер.

[@haliltokmakmatgeo]

💫💫💥

[@9WEAVER9]

GOODNESS

[@pranavamali05]

Thnx

[@notlin1976]

The 2nd solution is top! 🇧🇷🇧🇷🇧🇷🇧🇷🇧🇷🇧🇷🇧🇷🇧🇷

[@sachs6]

Just pretend the two equations had 5 in the base. Fake x and y have sum equal to 3 and difference equal to 2, so x/y is the ratio between fake 5/2 and fake 1/2. Fake and fake cancel out, as fake is the exponent that transformed 2 in 5 in the first place.

[@堀勇作-l5p]

second method is best. I hate log

[@devondevon4366]

2^5=32

[@yakupbuyankara5903]

32.

[@vatsal941]

32

[@mathswan1607]

32

[@shrikrishnagokhale3557]

32