Math Olympiad|Can you solve for x?| Algebra.

[@RanjithSenarathna-l1j]

🎉🎉🎉🎉

[@Futuristic-20]

Excellent!

[@godfreymabudo1765]

JJonlinemath. I am your favorite fan. I did maths decades ago but I always check RUclips to see your latest post. I love them all. Keep making me young. Thank you and Love from Uganda.
PS: By the way... where are you from?

[@francisamewode233]

Perfect. Love your teaching my love.

[@JJONLINEMATHSCLASSchannel]

Thanks

[@RyanLewis-Johnson-wq6xs]

3^x*5^x^2=15 x=1 x=-(Log[5,3]+1) Final answer

[@key_board_x]

3^(x) * 5^(x²) = 15
Ln[3^(x) * 5^(x²)] = Ln(15)
Ln[3^(x)] + Ln[5^(x²)] = Ln(15)
x.Ln(3) + x².Ln(5) = Ln(15)
x².Ln(5) + x.Ln(3) - Ln(15) = 0
Δ = [Ln(3)]² - 4.[Ln(5) * - Ln(15)]
Δ = [Ln(3)]² + 4.Ln(5).Ln(15)
Δ = [Ln(3)]² + 4.Ln(5).Ln(3 * 5)
Δ = [Ln(3)]² + 4.Ln(5).[Ln(3) + Ln(5)]
Δ = [Ln(3)]² + 4.Ln(5).Ln(3) + 4.Ln(5).Ln(5)
Δ = [Ln(3)]² + 2.[Ln(3) * 2.Ln(5) + [2.Ln(5)]²
Δ = [Ln(3) + 2.Ln(5)]²
x = { - Ln(3) ± [Ln(3) + 2.Ln(5)] } / 2.Ln(5)
First case: x = { - Ln(3) + [Ln(3) + 2.Ln(5)] } / 2.Ln(5)
x = [- Ln(3) + Ln(3) + 2.Ln(5)] / 2.Ln(5)
x = [2.Ln(5)] / 2.Ln(5)
→ x = 1
Second case: x = { - Ln(3) - [Ln(3) + 2.Ln(5)] } / 2.Ln(5)
x = [- Ln(3) - Ln(3) - 2.Ln(5)] / 2.Ln(5)
x = [- 2.Ln(3) - 2.Ln(5)] / 2.Ln(5)
x = - [Ln(3) + Ln(5)] / Ln(5)
→ x = - Ln(15) / Ln(5)

[@anestismoutafidis4575]

3^x•5^x^2=15 3•5=15
3^1 •5^1^2=15 3•5=15
3^ln(e) • 5^ln(e)=15^ln(e)
x={1; ln(e); sin(90)}

[@ntendeleni]

x=(-log5-log3/log5) not (-1-log3)/log5

[@abrahammelkonian9698]

OK !! X or sex ?

[@ubahprecious6151]

Diabolical😂