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.

@brandonhicks7549

4 күн бұрын

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.

@pavfrang

4 күн бұрын

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.

@jjeanniton

4 күн бұрын

Ergo: x = ln(5)/ln(2)

@pavfrang

4 күн бұрын

@@jjeanniton why using a base other than 2? It is clearly a base 2 logarithm: log2(5)

@AbbeyRoad69147

3 күн бұрын

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

@maxhagenauer24

2 күн бұрын

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.

@BubaSillah-uk6kw

2 күн бұрын

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) __________________ |x = ln(26)/ln(2)| __________________ |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 :-)

@alphalunamare

Күн бұрын

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. :-)

@alphalunamare

Күн бұрын

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