A Homemade Exponential Equation Solved in Three Ways

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PLAYLISTS 🎵 :

[WahranRai]

Raise both sides to power x*y and take log base 10 :
2^y = 5^x = 100^x*y ---> ylog2 = xlog5 = xylog100= 2xy ---> log2 = 2x and log5=2y ---> log2 + log5 = 2(x+y)---> x+y =1/2

[drdiegocolombo]

This is why I love this Mathematics. Multiple roads lead to the same destination.

[Jha-s-kitchen]

4:20 You said "Use Power of Properties" 😂😂🤣. Third method looked like an overkill at first, but the end result was very satisfying!

[mathswan]

1/2
2=100^x
5=100^y
Multiplying together
10=100^(x+y)=10^2(x+y)
1=2(x+y)
x+y=1/2

[ulrichkaiser]

nth method: I was simply using log base 10:
1/x * log2 = log100 = 2 ==> 2x = log2
1/y * log5 = log100 = 2 ==> 2y = log5
Adding: 2x + 2y = log2 + log5 = log(2*5) = log10 = 1
2x + 2y = 1 => 2(x+y) =1 ==> x+y = 1/2

[noeticresearch]

This is an excellent math problem that utilizes the properties of logarithms and natural logarithms Thank you SyberMath!

[gdmathguy]

I did a natural logarithm from the start of 5^(1/y)=100 and 2^(1/x)=100, then I multiplied by y and divided by ln(100), so y is ln(5)/ln(100) and x is ln(2)/ln(100). When adding them, it results in [ln(2)+ln(5)]/ln(100). Due to the product rule, ln(2)+ln(5)=ln(10) and due to the power rule, ln(100)=2*ln(10) and both ln(10) cancel out giving me 1/2. Very nice 👍

[GirishManjunathMusic]

Given:
2↑(1/x) = 100 — ①
5↑(1/y) = 100 — ②
To find:
x + y

As, by definition, x and y ≠ 0:

From ①:
2↑(1/x) = 100

Taking both sides to the power of x:
100↑x = 2 — ③

From ②:
5↑(1/y) = 100

Taking both sides to the power of y:
100↑y = 5 — ④

Multiplying ③ and ④:
(100↑x)·(100↑y) = 2·5 = 10

Using (a↑b)·(a↑c) = a↑(b + c):
100↑(x + y) = 10
(10↑2)↑(x + y) = 10

Using (a↑b)↑c = a↑(bc):
10↑2(x + y) = 10
10↑2(x + y) = 10¹

Taking decimal logarithm on both sides:
2(x + y) = 1
x + y = ½

[mcwulf]

The first method is less elegant but it's so quick to solve - took me a minute or so - I didn't bother trying to be any cleverer!!!

[pedrovargas]

first method is my instinctual go-to, second is the best.

[SuperYoonHo]

Awesome video sir. LOve it. That was a beautiful way of doing it. Thanks so mucH.
Oh, no!!! Your voice!!!

[blueray]

Great explanation, I gladly understood all the steps.

[Bjowolf]

Much easier just using the base 10 logarithm (log) on both sides of each equation:

2^(1/x) = 5^(1/y) = 100

log(2^(1/x)) = 1/x * log 2 = 2
log(5^(1/y)) = 1/y * log 5 = 2

x = log 2 /2
y = log 5 /2

x + y = (log 2 + log 5) / 2
= log(2*5)/2
= log 10 / 2
= 1/2

[matthewmanzanares]

I used the first method but I used natural log because it's a bit less confusing.

[MrLidless]

Why do YouTube creators always use logs with different bases for these questions? Ln, ln, ln!
1 / x = ln100 / ln2, 1 / y = ln100 / ln5
x+y = (ln2 + ln5) / ln100 = ln10 / ln100 = 1/2

[scottleung]

I used a fourth method, in which I solved for x and y individually and added them up.

[susmitadutta]

I solved it by taking a constant which is 'k'.

[tbg-brawlstars]

Is the answer 1/2 ?
I don't believe my answer because it was damn easy.
Edit : I am right
And my method was different lmao 😂

[morteza]

This issue is very simple. To solve it, we use the false root method, another one of my inventive methods🤠
2^(1/x)=5^(1/y)=100
2^(1/x)×5^(1/y)=10⁴=2⁴×5⁴
1/x=4 x=1/4 1/y=4 y=1/4
X+y=1/4+1/4=1/2😎
Next example
If 8^(a)=27^(b)=125^(c)=30
abc/(ab+bc+ac)=?
2^(3a)×3^(3b)×5^(3c)=30^(3)
2^(a)×3^(b)×5^(c)=2×3×5
a=1 b=1 c=1
abc/(ab+bc+ac)=1/3 easy
Next example
2^x=7^y=196 xy/(x+y)=?
2^(x)×7^(y)=(14)⁴=2⁴×7⁴
X=4 y=4 xy/(x+y)=2

[robertveith]

The third method is horribly convoluted, and its flow does not have an intuition about it. Thumbs-down for it.