Canada | A Nice Exponential Algebra Problem | Math Olympiad
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- Опубликовано: 9 сен 2024
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i have been done it
Is easy with me
To solve this equation, we need to first rewrite it by combining the exponents using the property a^(b^c) = a^(b*c):
2^x * 3^(x^2) = 6
Rewriting 3^(x^2) as (3^x)^2:
2^x * (3^x)^2 = 6
Now, rewrite 3^x as (2^1.5849625)^x and substitute it into the equation:
2^x * (2^1.5849625)^2x = 6
Now, apply the property of exponents that says if you have the same base, you can add the exponents:
2^x * 2^(3.169925) = 6
Adding the exponents:
2^(x + 3.169925) = 6
Since 2^2.5849625 ≈ 6.349604207, we get:
2^(x + 3.169925) ≈ 6.349604207
Now, take the log base 2 of both sides to find x:
x + 3.169925 = log2(6.349604207)
x + 3.169925 = 2.648461
x ≈ -0.521464
Therefore, the approximate value of x that satisfies the equation 2^x * 3^(x^2) = 6 is x ≈ -0.521464.
Not quite right.
3^(x^2), as shown in the problem, is the same as 3^(x * x), or (3^x)^x.
log(3^(x²)) = x² * log(3)
(3^x)^2, which you believe is the same as 3^(x^2), is actually 3^(x * 2), or 3^(2x).
log((3^x)^2) = 2 * log(3^x)
= 2x * log(3)
x² * log(3) is not the same as 2x * log(3).
Wrong already in the first line: a^(b^c) is _not_ the same as a^(b*c).
(a^b)^c = a^(b*c) would be right.
But (a^b)^c is _not_ the same as a^(b^c).
"Now, rewrite 3^x as (2^1.5849625)^x "
Why should we? That would result in an _approximate_ solution, with rounding errors. We are looking for _exact_ solutions here.
Hi guys,i don't know why but is seem ok to me
@@Nguyễn-j9q Huh? You still insist that your solution is right?
Amei!
Just simply put x= 1
And it will satisfy above equation 😅
Or comparing 2,3 powers
Такие ДОЛГИЕ ПРЕДВПРИТЕЛЬНЫЕ ЛАСКИ - а КОНЕЦ, всё равно "ОДИН"!
Решила устно. Х равен 1.
Not as easy as that, Fergusso!😝😝😝❌❌❌