A Fun Functional Equation with Exponents

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[@ameurzaheg5647]

There is a more elegant way to solve it in my opinion, my goal was to express 4^x in terms of 2^x +1:
f(2^x+1)
=4^x
=2^2x
=(2^x)^2 -1+1
=(2^x +1)(2^x -1)+1
=(2^x +1)(2^x +1-2)+1
Now we let 2^x+1=z, we have:
f(z)
=z(z-2) +1
=z^2-2z+1
=(z-1)^2
which means:
f(x)=(x-1)^2

[@vgautamkrishna5197]

F(x)= (x-1)²

[@robertibatullin8329]

The first method looks deliberately over-complicated. And the 2nd method can be presented even simpler:
Substitute y = 2^x. Get f(y+1) = y^2.
Substitute x = y+1. Get f(x) = (x-1)^2.

[@SidneiMV]

awesome!

[@RealQuInnMallory]

5 (x ➖ 5x+5)

[@allanmarder456]

The answer f(x)=(x-1)^2 is only partially correct. You also need to stipulate that x>1 to be consistent with the domain of the function given in the thumbnail.

[@d-8664]

Why though? x can be smaller than 1.

[@user-xz8vt2wf1g]

Πεδίο ορισμου(2,+00)

[@nizamkanchon3311]

Please help me solve the differential eqn.
y2+y1+2=0

[@maxhagenauer24]

It it's y^2 + y^1 + 2 = 0 then it's not even a differential equation, just a quadratic where y = ( -1 ± i*sqrt(7) ) / 2 but if it's y'' + y' + 2 = 0 then y = c_1 * e^(-x) + c_2 - 2x.

[@nizamkanchon3311]

@@maxhagenauer24 no brother, i meant y2 as second derivative of y and y1 as 1st derivative of y. Anyway, thnk u for trying to help.

[@DonEnsley-mathdrum]

problem
f(2ˣ+1) = 4ˣ
y=2ˣ+1
y-1=2ˣ
ln(y-1)=x ln 2
x = ln(y-1)/ln 2
f(y) = 4ˡⁿ⁽ʸ⁻¹⁾ᐟˡⁿ ²
= (eˡⁿ ⁴) ˡⁿ⁽ʸ⁻¹⁾ᐟˡⁿ ²
But ln 4/ ln 2 is just 2.
f(y) = (e²) ˡⁿ⁽ʸ⁻¹⁾
f(y) = (eˡⁿ⁽ʸ⁻¹⁾)²
= (y-1)²
answer
f(x) = (x-1)²

[@cosmolbfu67]

take log(base2) is easier