Canada | A Nice Exponential Algebra Problem | Math Olympiad

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[@Nguyễn-j9q]

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.

[@epsi]

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).

[@bjornfeuerbacher5514]

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).

[@bjornfeuerbacher5514]

"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.

[@Nguyễn-j9q]

Hi guys,i don't know why but is seem ok to me

[@bjornfeuerbacher5514]

@@Nguyễn-j9q Huh? You still insist that your solution is right?

[@soniamariadasilveira7003]

Amei!

[@divyakantojha6299]

Just simply put x= 1 And it will satisfy above equation 😅

[@divyakantojha6299]

Or comparing 2,3 powers

[@user-kd9oy5qn1y]

Такие ДОЛГИЕ ПРЕДВПРИТЕЛЬНЫЕ ЛАСКИ - а КОНЕЦ, всё равно "ОДИН"!

[@user-lx4uw4zn1u]

Решила устно. Х равен 1.

[@wira2562]

Not as easy as that, Fergusso!😝😝😝❌❌❌