Solving a Nice Exponential Equation from Romania

[@yoav613]

The only (real) solution for this problem.nice and easy😃

[@arbenkellici3808]

Very nice and easy solution professor

[@MaheshKumar-lx1ku]

Nice observation

[@broytingaravsol]

i reckoned the formula as 3²-7=2, so x=2

[@edcagas4893]

There is another solution I think

[@CriticSimon]

what is it?

[@walterwen2975]

Solving a Nice Exponential Equation from Romania:
3^[(x + 2)/(3x - 4)] - 7 = 2{3^[(5x - 10)/(3x - 4)]}; x = ?
3^[(x + 2)/(3x - 4)] > 7, No Complex value for x
Let: y = 3^[(x + 2)/(3x - 4)], 3^[(x + 2)/(3x - 4)] - 7 = y - 7 = 2{3^[(5x - 10)/(3x - 4)]}
(y - 7)y = 2{3^[(5x - 10)/(3x - 4)]}y = 2{3^[(5x - 10)/(3x - 4)]}{3^[(x + 2)/(3x - 4)]}
= 2{3^[(5x - 10 + x + 2)/(3x - 4)]} = 2{3^[(6x - 8)/(3x - 4)]} = 2(3^2) = 18
y^2 - 7y - 18 = 0, (y - 9)(y + 2) = 0, y + 2 > 0; y - 9 = 0
y = 3^[(x + 2)/(3x - 4)] = 9 = 3^2, (x + 2)/(3x - 4) = 2, x + 2 = 6x - 8, 5x = 10; x = 2
Answer check:
3^[(x + 2)/(3x - 4)] - 7 = 3^(4/2) - 7 = 2
2{3^[(5x - 10)/(3x - 4)]} = 2{3^[(10 - 10)/(6 - 4)] = 2(3^0) = 2; Confirmed
Final answer:
x = 2