A nice Math Olympiad Equation | Lambert w function Reveals Simplification | What is the Value of x ?

[@wes9627]

Substitute x=y^2 into the given equation, divide by 5, and rearrange to y=ln(27-y)/ln5. Inspection yields y=2 and x=y^2=4.
Check: 5^3+5√4=125+10=135

[@mustafapolat9014]

For the inspection you dont need ln. 5^y=27-y. y=2 is obviously solution.

[@walterwen2975]

A standard math solution:
135 > 5^(√x + 1) > 5√x > 0, 135 = 5(27) = 5(25 + 2) = 5^3 + 5(2)
5^(√x + 1) + 5√x = 135 = 5^3 + 5(2) = 5^(2 + 1) + 5(2) = 5^(√4 + 1) + 5(√4)
√x = √4; x = 4
Answer check:
5^(√x + 1) + 5√x = 135; Confirmed as shown
Final answer:
x = 4

[@pspprabhat]

=>5^_/×+1
=-5(_/×)+135
=>5^_/×=(27-_/×)
=>(1)=(27-_/×)5^-_/×
=>1)ln5(27)=(27-_/×).e^ln5(-×).ln5(27)
=>(5^2)ln5.e^ln5(25)
=(27-_/×)ln5.e^ln5(_/×).e^ln5(27)
=>W{25ln5.e^ln5(25)}
=W{(27-_/×)ln5.e^ln5(27-_/×)}
=>25ln5=(27-_/×)ln5
=>25=27-_/×
=>_/×=27-25=2
&×=4