A Math Olympiad exponential math problem| solve for x,y.

This is a tricky Math Olympiad exponential math question. #exponentialequationproblems #math

Always anticipate for my videos around 9 am and 9 pm WAT

[sylvesterogbolu-otutu]

x = 6 or 3
y = 3 or 6

x + y = 6 + 3 or 3 + 6 = 9.

Quick Working Method.

I will rewrite 15750 as:

15625 + 125.

Then, rewrite 15625 in exponential form as 5^6 and 125 as 5^3.

The entire expression becomes:

5^x + 5^y = 5^6 + 5^3

Equating exponents;

5^x = 5^6 or 5^x = 5^3

5^y = 5^6 or 5^y = 5^3

Therefore;

x = 6 or 3, and
y = 6 or 3.

Whatever the case;

x + y = 6 + 3 or 3 + 6 = 9.

[Psykolord]

Before watching:

Alright, so past the power of 2, every power of 5 (that I can find anyways) ends in either 125 or 625, and these alternate; odd powers result in numbers ending with 125, and even powers result in numbers ending with 625. Given that we must have one odd and one even power (otherwise x+y is not odd), this checks out.

I want to first check the closest power to our result. 5^3 = 125, 5^4 = 625, 5^5 = 3125m 5^6 = 15625, so let's look at 5^6.

15750-15625 = 125. Which...is just 5^3. And 3+6=9

Ergo (x, y) = (3, 6) or (6, 3) (order really doesn't matter here)

[vdkycrs]

thank you, it seems to me that the squaring of the equation is not necessary, if you write 15750 as the sum of 5 power 6 and 5 power 3, you get x and y by comparing the terms

[key_board_x]

5^(x) + 5^(y) = 15750 → given: x + y = 9 → y = 9 - x
5^(x) + 5^(9 - x) = 15750
5^(x) + [5^(9) * 5^(- x)] = 15750 → given: a = 5^(x) → where: a > 0
a + [5^(9) * (1/a)] = 15750
[a² + 5^(9)]/a = 15750
a² + 5^(9) = 15750a
a² - 15750a + 5^(9) = 0
Δ = (- 15750)² - [4 * 5^(9)] = 248062500 - 7812500 = 240250000 = 15500²
a = (15750 ± 15500)/2
a = 7875 ± 7750

First case: a = 7875 + 7750
a = 16625 → recall: a = 5^(x)
5^(x) = 16625
5^(x) = 5^(6)
→ x = 6
→ y = 3 ← because: y = 9 - x

Second case: a = 7875 - 7750
a = 125 → recall: a = 5^(x)
5^(x) = 125
5^(x) = 5^(3)
→ x = 3
→ y = 6 ← because: y = 9 - x

Other way without any indication about (x + y)
5^(x) + 5^(y) = 15750
5^(x) + 5^(y + x - x) = 15750
5^(x) + 5^(x + y - x) = 15750
5^(x) + [5^(x) * 5^(y - x)] = 15750
5^(x) * [1 + 5^(y - x)] = 15750 → 5^(x) is odd and 5^(y - x) is odd so, [1 + 5^(y - x)] is even
5^(x) * [1 + 5^(y - x)] = 125 * 126
5^(x) * [1 + 5^(y - x)] = 5^(3) * 126 → by identification

5^(x) = 5^(3)
→ x = 3

[1 + 5^(y - x)] = 126
5^(y - x) = 125
5^(y - x) = 5^(3)
y - x = 3
y = 3 + x → we 've just seen that: x = 3
→ y = 6

[RyanLewis-Johnson-wqxs]

5^x+5^y=15750 x+y=9 x=6 y=3 It’s in my head.