Can You Pass Cambridge Entrance Exam?

With Cardano/Tartaglia formula we get the following reasonning: 8^x + 2^x = 130 (2^3)^x + 2^x = 130 (2^x)^3 + 2^x = 130 get k = 2^x k^3 + k = 130 k^3 + k - 130 = 0 k^3 + k - 130 = 0 is based on k³ + kp + q = 0 template, with: • p = 1 • q = -130 recall and application of the Cardano/Tartaglia formula: • k = [-q/2 + √(q²/4 + p³/27)]^(1/3) + [-q/2 - √(q²/4 + p³/27)]^(1/3) k = (-(-130)/2 + √(((-130)^2)/4 + ((1)^3)/27))^(1/3) + (-(-130)/2 - √(((-130)^2)/4 + ((1)^3)/27))^(1/3) k = 5 2^x = k = 5 2^x = 5 x = ln(5)/ln(2) x = 2.321928 ----- check ----- 8^x + 2^x = 130 8^2.321928 + 2^2.321928 = 129.999997 ----- final results ----- ■ x = ln(5)/ln(2) ■ x = 2.321928 🙂

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.

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.

You could substitute y=2**x then use the rational root theorem to find the factor (x-5)

Interesting video, but if you're going to write the product as 26*5, why not just write it as 25*5 + 5 since you have y^3 + y, so 125+5 = y^3 + y, 5^3 + 5 = y^3 + y therefore y = 5 x = log_2(5).

If you don't spot a solution (x=5) the cubic can easily be solved by radicals.

8^x + 2^x = 130 2^3x + 2^x = 130, let U = 2^x u^3 + u = 130 I’m just gonna use my intuition at this point (if that wasn’t possible I’d just use newtons method). u = 5 so 2^x = 5 so x = log2(5) = ln(5)/ln(2)

Super 👍 from Algeria

Or we could just write it as: One hundred and dirty and solve it for y.

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

Log base 2 of 5 falls out by simple inspection! Why so difficult?

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?

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) ________________ |x = ln(26)/ln(2)| ________________ |x ≈ 4.70043... |

2^x = 5, then you put in log base 2 of x is 5

the trick is 130=26*5

BarryM is correct!!More inspiration, less perspiration!!

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

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 :-)