This equation has a problem because : 1225= 7^y =5^x and this is equal to 35^2 so (7x5)^2 . Then √(7^y) x √(5^x)= 7^2x5^2 so the only solution is y=4 and x=4 or, x≠y because 7^y=5^x . However, with my method we of course fall back to xy/x+y=0.5 I got stuck on it for an hour, please tell me if there is another solution than x=4=y.
@@PreMath This equation has a problem because : 1225= 7^y =5^x and this is equal to 35^2 so (7x5)^2 . Then √(7^y) x √(5^x)= 7^2x5^2 so the only solution is y=4 and x=4 or, x≠y because 7^y=5^x . However, with my method we of course fall back to xy/x+y=0.5 I got stuck on it for an hour, please tell me if there is another solution than x=4=y.
hi. Thanks for your problems, explanations and solutions. I solved this one using logarithms. x =ln (1225) / ln (5) ; y = ln (1225) / ln(7) . then i had to use 7*5 = 35 and 35² = 1225. my ending was xy / (x+y) = ln(1225)/ln 35) = 2 ln(35) / ln (35) = 2.
@@arekkrolak6320 Thats the way I did it too. The hardest thing was to find (5×7)²=1225 but you can guess it because this kind of question thend to fall apart and becone something simple
Sir,, actually in your channel one of the first things I like is comments,, bec different people expressing there different approaches and diff methods,,this makes me interesting ☺☺
I used logarithm! I first realized that the RECIPROCAL of xy/x+y is x+y/xy = 1/x + 1/y. x = lg 1225/lg 5. y = lg 1225/lg 7. 1/x + 1/y = lg 5/lg 1225 + lg 7/lg 1225 = lg 35/lg 1225. (Since 1225 = 35 x 35), 1/x + 1/y = lg 35/(lg 35^2) = lg 35 / (2 lg 35) = 1/2. Taking reciprocal of 1/2 to obtain answer to original equation = 2.
I have found another way to solve this. 1225=7^2•5^2 So knowing 5^x=7^y=1225, dividing first by 5^2 gives us 5^(x-2)=7^2 Put every number in this sequence to the y’th power and you get 5^y(x-2)=7^2y Now replace 7^y by 5^x as it is equal. We now are left with the equation 5^(yx-2y)=5^2x . Since the base is the same we have a^c=a^d and we know that c=d or in this case xy-2y=2x Now set the 2y to the other side and we have xy=2x+2y or xy=2(x+y) Thus we know that x+y is half xy so xy/x+y= 2 Keep up making these videos , greetings from Greece 🇬🇷
I took this way: (5^x) ^y=1225^y; (7^y)^x=1225^x, so we have 5^(xy)*7^(xy)=1225^(x+y) and finally 35^(xy)=35^[2*(x+y)]. Consequently, we obtain the result in easy way. Compliments for your work.
Ek ghante kosis ki magar nhi bana...lekin solution dekh ke crystal clear samajh me aaya🤗🤗 ...aur maine subscribe bhi kr diya aur amazing videos dekhne k liye
35=1225^{x+y/xy} can be written by x+y/xy = log base-1225 X 35 which gives 1225^x+y/xy=35 or 35^2(x+y/xy)= 35 such x+y/xy became 1/2 and it's reciprocal became 2
I think we can move to the final answer by noting that 1/x + 1/y is the same as ( x + y ) / ( x*y ) and that 1225 = 5^2 * 7^2 and the result follows quickly.
i use logarithm for this but i simplify that (log_5(1225)log_7(1225))/(log_5(1225)log_7(1225)) Since y = xlog_7(5) Then substitute to equation (log_5(1225)^2 log_7(5))/(log_5(1225)+log_5(1225)log_7(5)) Factor the denominator first (log_5(1225)^2 log_7(5))/log_5(1225)(1+log_7(5)) (log_5(1225)log_7(5))/(1+log_7(5)) log_5(1225) × log_7(5)/log_7(35) log_5(1225) × log_35(5) 2log_5(35) × log_35(5) 2 × log_35(35)/log_35(5) × log_35(5) 2 × 1 = 2 More harder if you use the logarithm to get the approximate value 😂
Let 5^x=7^y=1225 be equal to k K=5^x, 5=k^1/x, 7=k^1/y We can see that 1225=7²•5², by substituting the valus of 7 and 5 we get that 7²•5²=k^2/y•k^2/x and this is equal to 5^x or 7^y which is equal to k so we get:- K=K^2/y•K^2/x=K^2(x+y)/xy => k=k^2(x+y)/xy, hence 1/2=x+y/xy, so xy/x+y=2... Please pin this easy method🙏🙏
Do Bowling Pins Have Mathematics?? Comment Below about Your Idea P.S : If you Don't know the answer, watch this video about "Triangular Numbers" th-cam.com/video/KE_IKwgyEfM/w-d-xo.html
It can be solve simply by visualising that xy/x+y =1/(1/x+ 1/y). x= log 1225 base 7 and y = 1225 base 5. 1/x = log 7 base 1225 and 1/y= log 5 base 1225. Adding these two will yield= log 35 base 1125 = 1/2. Inversing it will give 2 as answer
Más claro y explicado, no se puede El idioma no me impide comprender el desarrollo del ejercicio profesor Gracias por aportar mas información a mi cerebro
I would use logarithms too. The solution presented in the video requires that we see ahead that 1225 is the square of 35 and that we could manipulate the given data to get a base of 35 on the other side of the equation. Using logarithms requires not intuition or 'seeing ahead.'
Me too. Taking the logarithm route, the expression simplified down to log(1225)/(log(5)+log(7)). Punching that into a scientific calculator, it equals 2. It's a good problem, not too difficult but it takes you back to fundamentals.
Yesssssssssss. I got it, very nice. I really like this question with a very simple answer Edit: I think that I solve in a more easy way. I used logs, so x = log_5 (1225) and y = log_7 (1225). 1225 = 5² × 7², so I use this to brake the logs for x and y using the log properties to get x = 2 + 2 log_5 (7) and y = 2 + 2 log_7 (5). After this, just put the x and y in the xy/(x+y), do the distribution for xy and you can simplify nicely at the end.
This is way easier. BUT This is for math olympics. There you are only allowed to use what you already had in school. If you use anything else you have to proof it. So I guess it was for 8 or 9 graders. Otherwise it would be way to easy for math olympics
I mean its very easy to understand once you know how to solve but i would never be able to do this. Its like he does some random algebra and then finds a solution
Since I felt 1225 lies between 5^5 and 5^4, I considered x as 4.5 and similarly for y as 3.5 approximately. Subs x as 4.5 n y as 3.5, I got the final answer as 2. Something like (9/2*7/2) /(9/2+7/2). Solved within 2 mins. Is it fine?
I made it different: 1225 =(7)²(5)². Then 5^x=(7)²(5)²=>5^x-2=7²=>x-2=(log5(7))2=>x=2(log7/log5+1). Same logic for Y. After calculate X and Y and change them in xy/(x+y), the answer is 2. Sorry if this is bad written, I speak spanish.
Hi. Thanx for this nice example to this general statement: Let a and b be positive real numbers, let n be a natural number and and let x be unequal -y. Then it always follows from the equation a^x = b^y = (a*b)^n that x*y / (x+y) = n. Proof: Same solution that PreMath showed us with a=5, b=7, n=2.
Yes, it generalizes. In fact, we have 5^x = 5²7², so x ln 5 = 2 ln 5 + 2 ln 7, and x = 2 (1 + ln 7/ln 5). To generalize, let r = ln 7/ln 5. Then x = 2(1 + r), and by symmetry, y = 2(1 + 1/r). Now, the pair X = (1 + r) and Y = (1 + 1/r) have the curious property that XY = X+Y, and thus XY/(X+Y) = 1, for any r ≠ 0, and this is independent of x, y, logs, exponents, and 1225. Hence, this could be worth memorizing for any prospective taker-of-future-Olympiads. Generalizing beyond this specific instance, if we had x = kX, y = kY for any k ≠ 0, then xy/(x+y) = k²XY/k(X+Y) = k. Finally, for this problem, we have k = 2.
Sir i was able to solve it by other method it is also easy You just need to equate everything to k then just write a equation in k And in the end by comparing powers you will get the same answer
By using logarithm , it can be solved easily! Answer is 2! Commenting before watching the video! I don't know what method is used in this video to solve this question.
Answer=2 I did it differently 1225= 35^2 5^x =35^2 (5^x)^1/x = (35^2)^1/x raised both to the power of 1/x 5 = 35^2/x 7^y= 35^2 (7^y)^1/y) =(35^2)^1/y raised both to the power of 1/y 7=35^2 5 x7 =(35^2/x)(35^2/y) =35^2/x+2/y multiply both equation and m^n x m^q= m^n+q 35^1 = 35^2/x +2/y 1 = 2/x +2/y equate the exponent 1 = 2(1/x +1/y) factor out 2 on the right side of the equation 1/2 = 1/x +1/y) divide both sides by 2 1/2 = (y +x)/xy add the right side of the equation xy =( x+y)2 xy/x+y =2 Answer if (n+m)/nm= 1/p then nm/(n+m)= p
I solved it like this first I rewrite 1225 as 5^2×7^2 then I multiplied 5 ^x and 7^x which gets equal to 5^4×7^4 then by comparing x, y = 4 then put values in question and we will get our result
Sir u can do it in another way... Here it is given that 5^x=7^y=1225... First we check that 1225=5^2*7^2... So we can write in this way:- 5^x*7^y= 1225*1225. i.e. 5^x*7^y= 5^4*7^4. i.e x=4, y=4 ( comparing the powers of both 5&7). Then we can write xy/(x+y)= 4*4/(4+4)= 2
1 , 4 , 9 , ____ Can You Guess the Next Number?? Comment Below P.S : If you Don't know the answer, watch this video about "SQUARE NUMBERS" th-cam.com/video/jg_kJ6MSDGk/w-d-xo.html
Dear Chandan, I don't know of any good book. However, if I were you I'd watch all the premath videos! That would give you a nice practice. Thanks for asking. I wish you all the best. You are awesome 😀 Love and prayers from the USA!
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Super explanation.
Thanks sir.
Glad to hear that!
Thank you for your feedback! Cheers!
You are awesome Govinda😀
Love and prayers from the USA!
@@HackedPC I don't think calculator is allowed for this question
@@PreMath sir what is wrong in these 👉👉 5^a×7^b=5^4×7^2 a=4 b=2
This equation has a problem because :
1225= 7^y =5^x and this is equal to 35^2 so (7x5)^2 . Then √(7^y) x √(5^x)= 7^2x5^2 so the only solution is y=4 and x=4 or, x≠y because 7^y=5^x . However, with my method we of course fall back to xy/x+y=0.5 I got stuck on it for an hour, please tell me if there is another solution than x=4=y.
@@PreMath This equation has a problem because :
1225= 7^y =5^x and this is equal to 35^2 so (7x5)^2 . Then √(7^y) x √(5^x)= 7^2x5^2 so the only solution is y=4 and x=4 or, x≠y because 7^y=5^x . However, with my method we of course fall back to xy/x+y=0.5 I got stuck on it for an hour, please tell me if there is another solution than x=4=y.
hi. Thanks for your problems, explanations and solutions. I solved this one using logarithms. x =ln (1225) / ln (5) ; y = ln (1225) / ln(7) . then i had to use 7*5 = 35 and 35² = 1225. my ending was xy / (x+y) = ln(1225)/ln 35) = 2 ln(35) / ln (35) = 2.
Same took no time with ln
Same here. Took me about 30 seconds to solve.
you can solve it even faster if you do not use common base as: xy/(x+y) = log5(1225)*log7(1225)/(log5(1225)+log7(1225)) :)
@@mario7501 good for you boy
@@arekkrolak6320 Thats the way I did it too. The hardest thing was to find (5×7)²=1225 but you can guess it because this kind of question thend to fall apart and becone something simple
You are doing an incredible service to all students around the world! Diligent. Patient. Caring.
So nice of you.
Thank you for your feedback! Cheers!
You are awesome Dr. Kelly 😀
Sir,, actually in your channel one of the first things I like is comments,, bec different people expressing there different approaches and diff methods,,this makes me interesting ☺☺
Sir, I try to solve this question about of 30 minutes but when i see your solution. I understand perfectly. Thank you Sir😁👍🙏
Welcome 👍
Excellent!
Thank you for your feedback! Cheers!
You are awesome Raghav 😀
@@PreMath please state if calculators are allowed
@@Jackrobert28
Hi Jack, in Olympiad questions, normally calculators are not permitted!
Thanks for asking. Keep it up 😀
Its 1 minute question
I used logarithm! I first realized that the RECIPROCAL of xy/x+y is x+y/xy = 1/x + 1/y.
x = lg 1225/lg 5. y = lg 1225/lg 7.
1/x + 1/y = lg 5/lg 1225 + lg 7/lg 1225 = lg 35/lg 1225.
(Since 1225 = 35 x 35), 1/x + 1/y = lg 35/(lg 35^2) = lg 35 / (2 lg 35) = 1/2.
Taking reciprocal of 1/2 to obtain answer to original equation = 2.
Excellent!
There are many ways to solve this problem.
Thank you for your feedback! Cheers!
You are awesome Garrick 😀
I have found another way to solve this.
1225=7^2•5^2
So knowing 5^x=7^y=1225, dividing first by 5^2 gives us 5^(x-2)=7^2
Put every number in this sequence to the y’th power and you get 5^y(x-2)=7^2y
Now replace 7^y by 5^x as it is equal. We now are left with the equation 5^(yx-2y)=5^2x . Since the base is the same we have a^c=a^d and we know that c=d or in this case xy-2y=2x
Now set the 2y to the other side and we have xy=2x+2y or xy=2(x+y)
Thus we know that x+y is half xy so xy/x+y= 2
Keep up making these videos , greetings from Greece 🇬🇷
Спосіб у відео кращий.
Excellent!
There are many ways to solve this problem.
Thank you for your feedback! Cheers!
You are awesome 😀
Love and prayers from the USA!
@@user-zx8dr1fo7p I know but this one is easier
You made it very complicated
Now look at this
Awesome👍👍👍
your solution is easy to follow and understand, thanks for sharing this Olympiad question
nice question sir thank u
Thanks very much.
Nice way to learn the power rule
Awesome
Behtreen koshan
Brilliantly explained. Awesome presentation . 👍👍
i have just multiplied the equation
5^x × 7^y = 1225×1225
5^x × 7^y = 35 ^4
5^x × 7^y = 5^4 ×7^4
x =4
y=4
xy/x+y
4×4/4+4=16/8 =2
I took this way: (5^x) ^y=1225^y; (7^y)^x=1225^x, so we have
5^(xy)*7^(xy)=1225^(x+y) and finally
35^(xy)=35^[2*(x+y)]. Consequently, we obtain the result in easy way.
Compliments for your work.
Woah
We used the same method independently from each other. Amazing
@@elias69420 Cool!
A Very nice math video!!
Great method prof
Need to know if it works by logarithms
Excellent exercise
Glad you think so!
Ek ghante kosis ki magar nhi bana...lekin solution dekh ke crystal clear samajh me aaya🤗🤗 ...aur maine subscribe bhi kr diya aur amazing videos dekhne k liye
very good, sir
Thank you so much, Sir
Nice one, thanks!
Nice , and easy problem
Thanks for liking
Excellent!
Glad to hear that!
You are awesome 😀
So nice 🌹🌹🇮🇳😃
Can you do a few more log questions please?
Your videos are very very useful! It's a great day when you upload!
Glad to hear that!
Thank you for your feedback! Cheers!
You are awesome Pratima 😀
Love and prayers from the USA!
Nice video
Thanks
Keep it up 👌
V nice
very well explained
Thanks sir
Most welcome
So nice of you.
Thank you for your feedback! Cheers!
You are awesome 😀
its very geniutic!
Thanks for video. Good luck!!!!!!!
Nice problem.
From-sri Lanka.
Thanks and welcome
So nice of you.
You are awesome AKD 😀
Love and prayers from the USA!
Is finding the possible values of x and y considered wrong since the answer to the question stops the math process?
another great question and solution - thank you
Very welcome David
Glad to hear that!
Thank you for your feedback! Cheers!
You are awesome 😀
Very neat solution,I made a mess of it,but your mathematic skill is very good, especially saying 35^2=1225,from this the solution is simpler.
Wonderful
Glad to hear that!
Thank you for your feedback! Cheers!
You are awesome Davis 😀
Can we use log
Excellent
Thank you for your feedback! Cheers!
You are awesome Susen 😀
Super problem. It's given in K.C.Nag in the chapter laws of indicies.
Nice
Super sir
I like you
35=1225^{x+y/xy} can be written by x+y/xy = log base-1225 X 35 which gives 1225^x+y/xy=35 or 35^2(x+y/xy)= 35 such x+y/xy became 1/2 and it's reciprocal became 2
I think we can move to the final answer by noting that 1/x + 1/y is the same as ( x + y ) / ( x*y ) and that 1225 = 5^2 * 7^2 and the result follows quickly.
Thanks for solution. You can also solve it by using logarithm.
Nice explanation, 🙂
Excellent!
Thank you for your feedback! Cheers!
You are awesome Mahalakshmi😀
i use logarithm for this but i simplify that
(log_5(1225)log_7(1225))/(log_5(1225)log_7(1225))
Since y = xlog_7(5)
Then substitute to equation
(log_5(1225)^2 log_7(5))/(log_5(1225)+log_5(1225)log_7(5))
Factor the denominator first
(log_5(1225)^2 log_7(5))/log_5(1225)(1+log_7(5))
(log_5(1225)log_7(5))/(1+log_7(5))
log_5(1225) × log_7(5)/log_7(35)
log_5(1225) × log_35(5)
2log_5(35) × log_35(5)
2 × log_35(35)/log_35(5) × log_35(5)
2 × 1 = 2
More harder if you use the logarithm to get the approximate value 😂
This was a fun one.
Normally, l got the value of x, y by the "ln" .thus put their in the equation.. Is That true?
Can we find x and y seperatly.....?
Why did you take the power of 5x and 1225 as 1/x?
I'm confused.
Thnx a lot
Welcome Pranav
Thank you for your feedback! Cheers!
You are awesome 😀
Very imp. Qus for ssc cgl 2021 ❣️😁😁😁🙏✌️
x log5=2log 35,
x=2log35/log5, & y=2log35/log7.
The given expression
1/{1/x +1/y}=log35 ÷ 2log35=2
Excellent!
There are many ways to solve this problem.
Thank you for your feedback! Cheers!
You are awesome 😀
that was cool 😎
Let 5^x=7^y=1225 be equal to k
K=5^x, 5=k^1/x, 7=k^1/y
We can see that 1225=7²•5², by substituting the valus of 7 and 5 we get that 7²•5²=k^2/y•k^2/x and this is equal to 5^x or 7^y which is equal to k so we get:-
K=K^2/y•K^2/x=K^2(x+y)/xy
=> k=k^2(x+y)/xy, hence 1/2=x+y/xy, so xy/x+y=2... Please pin this easy method🙏🙏
sử dụng logarit và casio cầm tay cũng ra nhé
I am a simple man. I was able to solve with logs. One day I will be clever like you.
Do Bowling Pins Have Mathematics??
Comment Below about Your Idea
P.S : If you Don't know the answer, watch this video
about "Triangular Numbers"
th-cam.com/video/KE_IKwgyEfM/w-d-xo.html
nice sir
Thanks and welcome
You are awesome Binyamin
Keep rocking😀
It can be solve simply by visualising that xy/x+y =1/(1/x+ 1/y). x= log 1225 base 7 and y = 1225 base 5. 1/x = log 7 base 1225 and 1/y= log 5 base 1225. Adding these two will yield= log 35 base 1125 = 1/2. Inversing it will give 2 as answer
5^x =1225 -(i) and 7^y =1225(ii)
multiply (i) and (ii)
5^X .7^y =1225^2
5^X .7^Y = 35^4
5^x .7^y = 5^4 .7^4
x=4 and y=4
xy/x+y =4.4/4+4 = 16/8 =2
The best explain 👍👍
Más claro y explicado, no se puede
El idioma no me impide comprender el desarrollo del ejercicio profesor
Gracias por aportar mas información a mi cerebro
¡Me alegra escuchar eso!
¡Gracias por tus comentarios! ¡Salud!
Eres increíble Luis 😀
¡Amor y oraciones desde EE. UU.!
Estoy de acuerdo contigo.
Great explanation!! It can also use ln to solve.
Yes, definitely!
Excellent!
Glad to hear that!
Thank you for your feedback! Cheers!
You are awesome 😀
@@PreMath I think the form is 5^x=7^y=5^n*7^n, and the answer will be n. It's funny.
Super problem , Sir.I did it by using logarithms and got the answer as 2.
Excellent!
There are many ways to solve this problem.
Thank you for your feedback! Cheers!
You are awesome Jaggi😀
I would use logarithms too. The solution presented in the video requires that we see ahead that 1225 is the square of 35 and that we could manipulate the given data to get a base of 35 on the other side of the equation. Using logarithms requires not intuition or 'seeing ahead.'
Me too. Taking the logarithm route, the expression simplified down to log(1225)/(log(5)+log(7)). Punching that into a scientific calculator, it equals 2. It's a good problem, not too difficult but it takes you back to fundamentals.
Yesssssssssss. I got it, very nice. I really like this question with a very simple answer
Edit: I think that I solve in a more easy way. I used logs, so x = log_5 (1225) and y = log_7 (1225). 1225 = 5² × 7², so I use this to brake the logs for x and y using the log properties to get x = 2 + 2 log_5 (7) and y = 2 + 2 log_7 (5). After this, just put the x and y in the xy/(x+y), do the distribution for xy and you can simplify nicely at the end.
This is way easier.
BUT
This is for math olympics. There you are only allowed to use what you already had in school. If you use anything else you have to proof it.
So I guess it was for 8 or 9 graders. Otherwise it would be way to easy for math olympics
@@PeterLE2 Hmm, I understand now. I simply thought in a easy way to do it, I don't thought in this type of limitation.
@@LuizFernando23250 Neither did I when I participated many many years ago 😃
I did it in same way under 3_4 minutes
@@PeterLE2 it was for 7 graders in my country
I mean its very easy to understand once you know how to solve but i would never be able to do this. Its like he does some random algebra and then finds a solution
Since I felt 1225 lies between 5^5 and 5^4, I considered x as 4.5 and similarly for y as 3.5 approximately. Subs x as 4.5 n y as 3.5, I got the final answer as 2. Something like (9/2*7/2) /(9/2+7/2). Solved within 2 mins. Is it fine?
💙🙏
I started with an invert. If z=?answer, then 1/z=1/x + 1/y= log 5/log 1225 +log7/log1225=(log5+log7)/log1225= log35/(2*log35)=1/2=>z=2
I made it different: 1225 =(7)²(5)². Then 5^x=(7)²(5)²=>5^x-2=7²=>x-2=(log5(7))2=>x=2(log7/log5+1). Same logic for Y. After calculate X and Y and change them in xy/(x+y), the answer is 2. Sorry if this is bad written, I speak spanish.
So easy
Hi. Thanx for this nice example to this general statement:
Let a and b be positive real numbers, let n be a natural number and and let x be unequal -y.
Then it always follows from the equation a^x = b^y = (a*b)^n
that x*y / (x+y) = n.
Proof: Same solution that PreMath showed us with a=5, b=7, n=2.
Yes, it generalizes. In fact, we have 5^x = 5²7², so x ln 5 = 2 ln 5 + 2 ln 7, and x = 2 (1 + ln 7/ln 5).
To generalize, let r = ln 7/ln 5. Then x = 2(1 + r), and by symmetry, y = 2(1 + 1/r).
Now, the pair X = (1 + r) and Y = (1 + 1/r) have the curious property that XY = X+Y, and thus XY/(X+Y) = 1, for any r ≠ 0, and this is independent of x, y, logs, exponents, and 1225.
Hence, this could be worth memorizing for any prospective taker-of-future-Olympiads.
Generalizing beyond this specific instance, if we had x = kX, y = kY for any k ≠ 0, then xy/(x+y) = k²XY/k(X+Y) = k.
Finally, for this problem, we have k = 2.
My guess is 2
Edit: yay I was right! My method was to solve for x and y graphically then plug in those values to the xy/x+y equation to get my answer.
just use lg on both sides for each case
Sir i was able to solve it by other method it is also easy
You just need to equate everything to k then just write a equation in k
And in the end by comparing powers you will get the same answer
So what power of 5 equals to 1225? Or what power of 7 equals 1225?
joli !
By using logarithm , it can be solved easily!
Answer is 2!
Commenting before watching the video! I don't know what method is used in this video to solve this question.
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Answer=2
I did it differently
1225= 35^2
5^x =35^2
(5^x)^1/x = (35^2)^1/x raised both to the power of 1/x
5 = 35^2/x
7^y= 35^2
(7^y)^1/y) =(35^2)^1/y raised both to the power of 1/y
7=35^2
5 x7 =(35^2/x)(35^2/y) =35^2/x+2/y multiply both equation and m^n x m^q= m^n+q
35^1 = 35^2/x +2/y
1 = 2/x +2/y equate the exponent
1 = 2(1/x +1/y) factor out 2 on the right side of the equation
1/2 = 1/x +1/y) divide both sides by 2
1/2 = (y +x)/xy add the right side of the equation
xy =( x+y)2
xy/x+y =2 Answer if (n+m)/nm= 1/p then nm/(n+m)= p
Hi. There is an easier way. 1225 is the 35². And 35 is 7*5. 5^x*7^y=1225^2. So basically you get that both x and y are equal to 4. Then xy/x+y=16/8=2
Same method i have applied
I solved it like this first I rewrite 1225 as 5^2×7^2 then I multiplied 5 ^x and 7^x which gets equal to 5^4×7^4 then by comparing x, y = 4 then put values in question and we will get our result
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"Mathematics is greatest magic of all times" 😍🔥🔥🔥😘🔥🔥😘🔥🔥🔥😍😍
we can also solve it by using 'log' method
Sir u can do it in another way...
Here it is given that 5^x=7^y=1225...
First we check that 1225=5^2*7^2...
So we can write in this way:- 5^x*7^y= 1225*1225.
i.e. 5^x*7^y= 5^4*7^4.
i.e x=4, y=4 ( comparing the powers of both 5&7).
Then we can write xy/(x+y)= 4*4/(4+4)= 2
Its simple
It's simple
3:11hey plz answer quickly plz I got an exam tomorrow.
Why can you multiply5 and7 ,1225 1/x and 1225 1/y
I did by prime factorizing 1225=5×5×7×7
30s and done. Easy one
Op sir
1 , 4 , 9 , ____
Can You Guess the Next Number??
Comment Below
P.S : If you Don't know the answer, watch this video
about "SQUARE NUMBERS"
th-cam.com/video/jg_kJ6MSDGk/w-d-xo.html
The solution is indeed excellent but how would I know by heart that 35^2=1225 🤔
Since I didn't find the value of x and y individual , I couldn't understand how 5x =7y are = 1225
I study in graduation and I want to appear the examination of math olympiad in India ,but I have no idea of olympiad examination ,please suggest me
Please anyone or premath jii
Dear Chandan, I don't know of any good book. However, if I were you I'd watch all the premath videos! That would give you a nice practice.
Thanks for asking. I wish you all the best.
You are awesome 😀
Love and prayers from the USA!
I solve it from different approach in almost 2min.😊
Applying logarithmic gets problem more easy
i.e, loga^b. = bloga