Solve for X challenge! | e^x-e^-x=2 | challenging Algebra Problem | Olympiad Math

In this video, we will solve for x in this nice exponential equation e^x-e^-x=2. we will solve this algebraic equation problem trending 2022. Math Olympiad problems
we will solve for all possible values of X. This is actually one of the challenging math problems so do not ignore this video.

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[SuperYoonHo]

Awesome you rock dude keep it up my freind1

[suryanshubansal]

Damn olympiads cant lower their level more..

[GodbornNoven]

not challenging.
It's very easy in fact
e^x - e^-x = 2
let e^x = a
e^-x = e^(-1 * x) = (e^x)^-1
a-a^-1 =2
a - 1/a = 2
multiply both sides by a
a²-1=2a
add two to Both sides and subtract 2a from both sides
a²-2a+1=2
(a-1)²=2
a-1=±r2
e^x=a=1±r2
to solve for x, we simply have to ln both sides, but-
1-r2 < 0
And logarithms aren't defined for values lower than 0, thus
x=ln(1+r2)

[suryanshubansal]

Bro this is so simple. My teacher gave it to me in my first quadratic equations class to solve by myself..

[markusbanach-stb]

Why this abc-thing? When a=1 just use the p, q-formular.