Math Olympiad question: 3ˣ = x⁹; x =? 3ˣ > 0, x > 0; (x⁹)¹⸍⁹ˣ = (3ˣ)¹⸍⁹ˣ, x¹⸍ˣ = 3¹⸍⁹ = (3³)¹⸍²⁷ = 27¹⸍²⁷; x = 27 Trial-and-error math to find more roots, if available: x = 1: 3ˣ = 3 > x⁹ = 1; x = 2: 3² = 9 < 2⁹, 2 > x > 1, Very close to 1 x = 1.1: 3¹·¹ = 3.348 < 1.1⁹ = 2.358 x = 1.15: 3¹·¹⁵ = 3.537 > 1.15⁹ = 3.518, 2 > x > 1.15 x = 1.151: 3¹·¹⁵¹ = 3.541 < 1.151⁹ = 3.546, 1.151 > x > 1.15, Slightly < 1.151 x = 1.1508: 3¹·¹⁵⁰⁸ = 3.540 = 1.1508⁹ = 3.540; No more root The calculation was achieved on a smartphone with a standard calculator app Answer check: x = 27: 3ˣ = 3²⁷ = (3³)⁹ = 27⁹ = x⁹; Confirmed x = 1.1508: 3¹·¹⁵⁰⁸ = 3.540 = 1.1508⁹; Confirmed as shown Final answer: x = 27 or x = 1.1508
Since x is at power 9, there may be up to 9 solutions ;) A second real solution is already listed in the comments, so the solving in the video is bad. DISLIKE.
Assuming there could be 9 solutions is incorrect. For example, 2^x = x^2 has 3 real solutions. x = -0.76666, 2, and 4. The order of x does not dictate the number of solutions.
@@BartBuzz You got the idea. There might be more than 2 solutions. A really good problem would how to demonstrate there are only 2 solutions. Beyond my skills.
@@SoreInMusic There are only 2 REAL solutions. If there are any complex solutions, that's beyond me too. It's not obvious how to prove the number of total solutions. The second problem in x^2 had 3 solutions. I don't know if there are more than 3. I only know that there are only 3 REAL solutions.
Cool! Going to look up the Lambert W function now!
Very nice Algebra practice.
This one was posted 4 months ago. I posted a LambertW solution in the comments. Thanks for all these questions. Keeps the brain going....
Math Olympiad question: 3ˣ = x⁹; x =?
3ˣ > 0, x > 0; (x⁹)¹⸍⁹ˣ = (3ˣ)¹⸍⁹ˣ, x¹⸍ˣ = 3¹⸍⁹ = (3³)¹⸍²⁷ = 27¹⸍²⁷; x = 27
Trial-and-error math to find more roots, if available:
x = 1: 3ˣ = 3 > x⁹ = 1; x = 2: 3² = 9 < 2⁹, 2 > x > 1, Very close to 1
x = 1.1: 3¹·¹ = 3.348 < 1.1⁹ = 2.358
x = 1.15: 3¹·¹⁵ = 3.537 > 1.15⁹ = 3.518, 2 > x > 1.15
x = 1.151: 3¹·¹⁵¹ = 3.541 < 1.151⁹ = 3.546, 1.151 > x > 1.15, Slightly < 1.151
x = 1.1508: 3¹·¹⁵⁰⁸ = 3.540 = 1.1508⁹ = 3.540; No more root
The calculation was achieved on a smartphone with a standard calculator app
Answer check:
x = 27: 3ˣ = 3²⁷ = (3³)⁹ = 27⁹ = x⁹; Confirmed
x = 1.1508: 3¹·¹⁵⁰⁸ = 3.540 = 1.1508⁹; Confirmed as shown
Final answer:
x = 27 or x = 1.1508
X=27 (may be
3^x=x^9
3=x^(9/x)
= (27)^(9/27)
=(27)^(1/3)
=3
Omitted few steps
After solving x=27
How many solutions has this equation?
x^3^2 (x ➖ 3x+3).
Is it the answer for the second?
@@Technique-r8l yes
case 1 , x*ln3=9*lnx , --> , ln3/9=lnx*e^(-lnx) , *(-1) , -ln3/9= -lnx*e^(-lnx) , -lnx=W(-ln3/9) , x=e^W(-ln3/9) , x=~ 1.15082 ,
test , 3^x=3.54065 , 9^x=3.54065 , OK ,
case 2 , -ln3/9=-3*ln3/27 , -3*ln3/27=-3*ln3/3^3 , -3*ln3=--3*ln3*e^-3ln3 , -lnx=--3ln3 , lnx=3ln3 ,
x=e^(3*ln3) , x=27 , test , 3^27=27^9 , 3^27=(3^3 )^9 , 3^(3*9)=3^27 , 3^27=3^27 , Ok , solu , x= 1.15082 , 27 ,
how did you go from w(-ln3/9)=-lnx to e^(w(-ln3/9))=x, would you not have to make it e^-(w(-ln3/9))=x? where does the negative go?
@@dah_bard1160 W(-ln3/9) =~ -0.140478921 , -lnx=-0.140478921 , lnx=0.140478921 , x=e^(0.140478921) ,
that's right : e^(-W(-ln3/9)) , thank you for your comment
х=1,1508
45
=0
You are fast
i cant believe it i did it in my mind in like 2 minutes
And…
@@anderslarsson7426 i am happy man i am typically that fast why be rude
Since x is at power 9, there may be up to 9 solutions ;) A second real solution is already listed in the comments, so the solving in the video is bad. DISLIKE.
Assuming there could be 9 solutions is incorrect. For example, 2^x = x^2 has 3 real solutions. x = -0.76666, 2, and 4. The order of x does not dictate the number of solutions.
@@BartBuzz You got the idea. There might be more than 2 solutions. A really good problem would how to demonstrate there are only 2 solutions. Beyond my skills.
@@SoreInMusic There are only 2 REAL solutions. If there are any complex solutions, that's beyond me too. It's not obvious how to prove the number of total solutions. The second problem in x^2 had 3 solutions. I don't know if there are more than 3. I only know that there are only 3 REAL solutions.
3ˣ = x⁹ > 0; x > 0
x = 28: 3ˣ = 3²⁸ = 2.29(10¹³) > x⁹ = 28⁹ = 1.06(10¹³)
x = 30: 3³⁰ = 2.06(10¹⁴) > 30⁹ = 1.97(10¹³)
When: x > 27, 3ˣ > x⁹; The two curves never meet again.