Can You Pass Cambridge Entrance Exam?
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- čas přidán 25. 06. 2024
- Entrance examination and Math Olympiad Question in 2020. If you're reading this ❤️.
What do you think about this problem?
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You could substitute y=2**x then use the rational root theorem to find the factor (x-5)
Once we had y^3 + y - 130 = 0, it was immediately clear that y = 5 and the problem was solved. I’m not sure that the approach of changing y to 26y - 25y is an easier approach; it seems it requires more insight. Your thoughts please.
There are two more solutions, though complex. I agree that if you can see the (y-5) factor, it’s probably faster to just do the poly division.
You divide the polynomial with y-5 and check the remainder, which is y²+5y+126. It's clearly a polynomial division issue. You should check the roots of the remainder because they could be real numbers too. So in general a solution other than 5 could be also feasible.
Ergo: x = ln(5)/ln(2)
@@jjeanniton why using a base other than 2? It is clearly a base 2 logarithm: log2(5)
Yep
As soon as I saw that we had a cubic to solve, and 130 has a prime factorisation to 3 primes, 2x5x13, I knew one of them was going to be a candidate! Also if you dont do the 26-25 trick, and consider what the coeficients of general cubic would be, given there is no y^2 coeficient, you get the property that the sum of the roots is 0, and the product of the roots is 26.
Super 👍 from Algeria
Or we could just write it as: One hundred and dirty and solve it for y.
Log base 2 of 5 falls out by simple inspection! Why so difficult?
After substituting y, we could factor out y and find the product of 130.
y(y^2+1)=130
y(y^2+1)=26×5
y=5
y^2+1=26
y^2=26-1
y^2=25
y=+5,-5 but y=5 is the same as th other one.
Therefore since both y are equal then y=5
Then we can solve for x
But you cant just turn 130 into any 2 numbers you want to then set y equal to them multiplied together because there are infinitely many combinations. You could make it 2*65 if you wanted and get something completely different.
It is about choosing a the product that captures it the best
2^x + 8^x = 130
2^x + 4 (8^x) = 130
2^x (1+4) = 130
2^x × 5 = 130
2^x = 130 ÷ 5
2^x = 26
xln(2) = ln(26)
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|x = ln(26)/ln(2)|
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|x ≈ 4.70043... |
u = 2^x
u^3 + u - 130 = 0
5 is the obvious real answer.
(u^3 + 0u^2 + u - 130)/(u - 5) = u^2 + 5x + 26. Quadratic equation for the two complex answers.
Then back substitute.
The real answer is easy: ln(5)/ln(2). Back substituting the complex answers are a pain in the butt and I'm not going to bother.
Maybe Cambridge High School?
BarryM is correct!!More inspiration, less perspiration!!
9^2+7^2=130
OK Space Cadets ..this should be easy 2^x + 4^x + 8^x + 16^x = 780 ........................ hint: the 3^x + 9^x + 27^x + 81^x = 780 results in x = Log to the base 3 of 5 :-)
We have : if n is integer greater than 1, then n^x + n^2x + n^3x + n^4x = 780 => x = Log to the base n of 5. Play with windows scientific calculator, have fun and generalise this stuff. The games are endless. :-) For example, why 5? choose any integer m and calculate m^3 + m. Suppose m = 7 and suppose n^x +n^3x = 350 then this results in x = Log to the base n of 7. In general Log n of m. This is the real fun, not the trickery of 26y - 25y. :-)
Here is the kicker!!!!!!! choose n = 7 of m = 9 and you get 738 as the result of the last equation. Not many people know this but there are 738 episodes of Startrek! So log to the base 7 of 9 is quite important ! How is that for contemporary magic?
Yeah, I did it, but not elegantly like you.
Up to substitution, same.
Then I got y(y^2+1)-130=0,
concluded that the bracket cannot be negative, and thus y must be a positive number.
Then I tried numbers like a scrub and ended up with y=5, thus after resub, log2(5), and that was good enough for me lol