Working with an Exponential Expression in Three Ways

If you need to post a picture of your solution or idea:
#ChallengingMathProblems #ExponentialExpressions #ExponentialEquations
via @YouTube @Apple @Desmos @NotabilityApp
@googledocs @canva

PLAYLISTS 🎵 :

[f-th]

Yeah the third method is probably the intended solution if notice early that the required expression looks similar to 1/x + 1/y = (x+y)/(xy)

[timeonly]

Did a slight variation of method 3:
2^x = 196 => 2^(xy) = 196^y
7^y = 196 => 7^(xy) = 196^x
Multiplying two equations:
14^(xy) = 196^(x+y) = (14^2)^(x+y) = 14^[2(x+y)]
So, equating exponents of the same base 14:
xy = 2(x+y) => xy/(x+y) = 2. Done!

[philstubblefield]

One of my favorites so far! Love the three different methods, none of which are terribly complicated, but all of which display exponent and logarithm rules so clearly.

[GourangaPL]

your 1st method was the one i came up with while solving this, most natural way to deal with such problem for me

[jmiki]

I used a little different approach, which kinda looks like an amalgamation of the first and the third method.

First, using logarithms to find the values of x and y (and then leave them that way):
x = log₂196
y = log₇196

Second, notice that in the wanted expression, there's a product in the numerator and a sum in the denominator which is usually preferred the other way around, so take the reciprocal:
xy/(x+y) = 1 / [(x+y)/xy] = 1 / (x/xy + y/xy) = 1 / (1/y + 1/x)

Finally, replace x and y with their values and use heavily the fact that for the reciprocal of the logarithm you can just switch the base and the argument (and other logarithmic identities):
1 / (1/log₇196 + 1/log₂196) = 1 / (log₁₉₆7 + log₁₉₆2) = 1 / log₁₉₆14 = log₁₄196 = log₁₄(14²) = 2

[aviratnakumar]

Any questions of this type has answer 2 of constant term is a perfect square and a multiple of both the variables.

If a^x= b^y=(ab) ^2 then (xy) /(x+y) =2

[tontonbeber]

Too simple.
Base of log doesn't matter
x = ln 196 / ln 2
y = ln 196 / ln 7
xy / (x+y) = [ (ln 196 . ln 196) / ( ln 2 . ln 7 ) ] / [ ln 196 ( ln 2 + ln 7 ) / ( ln 2 . ln 7) ]
= (simplification) = ln 196 / ( ln 2 + ln 7) = ln 196 / ln 14 = 2 ln 14 / ln 14 = 2

[WahranRai]

Let 14 the base of logarithm (seems appropriate to me : 2*7= 14 and 196 = 14^2)
xlog2= ylog7=log14^2=2 and log14= log2 + log7 = 1 ---->
1/x + 1/y = log2 /2 + log7 /2 = ( log2 + log7 )/ 2 = 1/2 = (x+y) /xy *xy/(x+y) = 2*

[morteza]

Another method is just to complete the methods, however, the best methods were told in the video, especially the third method.
xy/(x+y)=u
xy=ux+uy
x(y-u)=uy
X=uy/(y-u)
According to 2^(x)=7^(y)=(14)^2
We have 2^(uy/(y-u))=7^(y)
2^(u/(y-u))=7 u/(y-u)=log(7)/(log(2))
u/y=log(7)/(log(14))
u=log(7^(y))/(log(14)
u=log(14²)/log(14)=2

[bekaluu]

I used the first method but I really liked the second and third methods. Thanks a lot. Always enjoy your selections.

[pavelgatnar]

I raised 1st eq by y and 2nd eq by x:
2^xy=14^2y, 7^xy=14^2x
By multiplication:
14^xy=14^2(x+y)
Raise by 1/(x+y) and you have the result.

[tapassikder]

The last one was just a Magical Way. I loved it.

[Blaqjaqshellaq]

Another approach is with natural logarithms.
Therefore [xy/(x+y)]= [2*(ln2+ln7)/ln2 *2* which equals two.

[wuhyltw]

I think that the 1st method of using logarithm is the usual solution, but if you notice 2*7=14 and 196=14², it will be a simple and easy-to-understand 3rd method to solve by having equal exponential parts, even if there are some differences.

[ujjwalgupta]

I don't know why anyone hasn't thought of this but just multiply 2^x and 7^x together and we get (2^x)(7^x) = (2^4)(7^4). Then after comparing we get x=y=4. Plugging this into the expression we get 16/8 = 2

[silviatotaro]

To make a long story short...you are amazing!
Ciao, from Italy!

The harmonic mean of 3 numbers are given by 3xyz/(xy+xz+yz). For example if x=3, y=4, y=5, the formula gives 180/47. To show that this formula is correct consider the pairwise harmonic means: h(3, 4)=24/7, h(3, 5)=15/4, h(4, 5)=40/9. Then use these. again. This can of course be proven algebraically for any three non-negative numbers.

[Muslim_]

I liked this example and the ways you solved it 👍

In general: If a^x = b^y = ab, then the harmonic mean of x and y, h(x, y) = 2. And if a^x = b^y = c^z = abc, then h(x, y, z) = 3, and so on. In this case a^x = b^y =(ab)^2. xy/(x+y) is half the harmonic mean. 4/2 = 2.

[moeberry]

Too easy I did it in my head, but once again great video.