A Nice System Of Logs

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[@arpit_youtube]

take care sir.
god will make you well soon.

[@Nothingx303]

I hope your laryngitis will be fixed soon 😊😊😊

[@SyberMath]

Thank you!

[@damiennortier8942]

I hope you'll feel good very soon 😊
I liked more when you spoke.
Take care 🙂

[@SyberMath]

😊 thank you

[@alipourzand6499]

Very nice! 2nd method alternative:
x = 2^k, y = 2^(6-k), and wlog x and y are interchangeable.
Get well soon!

[@BalrajSingh-k5g]

By first approach .. thank you

[@SyberMath]

All the best

[@giuseppemalaguti435]

t=log_x(y)..ottengo due soluzioni 1..t=5..log_x(64/x)=5..(x=2)...2...t=1/5..log_x(64)=6/5..log_64(x)=5/6..x=64^(5/6)

[@prollysine]

log(x)/log(y)=5 , y^5=x , y^5*y=64 , y^6=2^6 , y=2 , x=64/2 , x=32 ,

[@scottleung9587]

Nice!

[@Skank_and_Gutterboy]

No problem, your health comes first.

[@SyberMath]

Thank you! You're awesome 🥰

[@rakenzarnsworld2]

x = 32, y = 2 or x = 2, y = 32

[@devondevon4366]

Answer 2 and 32
a different approach
log yx + xy = 26/5
This is akin to saying y is raised to a fraction of 26/5 = x, and x is raised to the remaining
fraction =y
let p =fraction of 26/5 which y is raised to or y^p=x Equation A
Hence, 26/5-p is the remaining portion that x is raised to or x^ 26/5 -p =y Equation B
y = x^ 1/p (raised both sides of equation A to 1/p) Equation C
x^ 1/p = x^ 26/5- p (transitive property since both = y)
1/p = 26/5 - p (relating the base)
1/p + p = 5 + 1/5
Hence p = 5
Hence, y^5 = x (substitute p=5 into Equation A)
Since xy= 64 (given), then x = 64/y
Hence y^5= 64/y
Hence, y^ 6 = 64
Hence y =2
Hence, x = 32 (2^5)

[@leonardobaraviera1759]

Logaryngitis!!!!

[@SyberMath]

😜😍

[@MisterPenguin42]

Bro, just AI your voice! Good music though. Hate to go a day without "Happy birthday 2 U"

[@Don-Ensley]

problem
log ᵧ x + log ₓ y = 26 / 5
xy = 64
Convert to natural logarithms.
ln x / ln y + ln y / ln x = 26/5
Let
u = ln x / ln y
u + 1/u = 26/5
5u + 5/u = 26
5 u²- 26 u + 5 = 0
Quadratic formula
u = [26 ± √(26²-10²)] /10
= {26 ± √[(26-10)(26+10)} /10
= (26 ± 24)/10
= 1/5, 5
For u = 1/5:
ln x / ln y = 1/5
5 ln x = ln y
e^(5 ln x) = y
y = x⁵
xy = 64
= x⁶
x = ± 2
y = ± 32
For u = 5:
ln x / ln y = 5
5 ln y = ln x
e^(5 ln y) = x
x = y⁵
xy = 64
= y⁶
x = ± 32
y = ± 2
We always have xy positive 64. Opposite signs are not in the solution set.
answer
(x,y) ∈ { (-32,-2),(-2,-32),(2,32),(32,2) }
It has been noted that the negative values must be excluded because the arguments to logs can't be negative. So the answer should be
(x,y) ∈ { (2,32),(32,2) }

[@Skank_and_Gutterboy]

Negative numbers are not part of the solution set because a logarithm cannot have a negative base. Other than that, I did this just like you did.

[@Don-Ensley]

@@Skank_and_Gutterboy
(-2)⁵ = -32
same as
log ₋₂ (-32) = 5

[@Don-Ensley]

@@Skank_and_Gutterboy i have corrected. 🙏