A Nice Exponential Equation | An Algebra Challenge

[@mohammedsaysrashid3587]

It was wonderful explanation thanks for sharing Sir 🙏....x=2/3

[@RashmiRay-c1y]

The equation can be written as 2^(3x) + 4 = 1/8 2^9x. Let t = 2^3x > t^3= 8t+32 > t=4 is a solution. [t^3-8t-32]/(t-4) = t^2+4t+8. t^2+4t+8 = 0 has the solutions t = -2 +/- 2 i which are complex and not acceptable. Thus t=4 > x = 2/3.

[@dorkmania]

Simplifying RHS to (2^3x)/2 and substituting for LHS radical with t³ = (2^3x) + 4, the equation becomes
(t³)^⅓ = (t³ - 4)/2
=> 2t = t³ - 4
=> t³ - 2t - 4 = 0
Using RRT and SDM, we get t = 2 as the only real solution.
2^3x - 4 = (2)³
=> x = ⅔

[@woobjun2582]

The given can be
³√(t +4) = t/2
by letting t = 2³ˣ
Cubing and rearringing,
t³-8t -32 = 0;
t³ -64 -8t +32 = 0;
(t³ -4³) -8(t -4) = 0;
(t -4)(t² +4t +16) -8(t -4) =0;
(t -4)(t² +4t +8) =0
That is,
t -4 =0 ; t = 4
or
t² +4t +8 = 0; t = cmplx
Therefore,
t = 2³ˣ = 4 = 2²
Finally
3x =2; x = 2/3

[@walterwen2975]

An Algebra Challenge: ³√(8ˣ + 4) = √4³ˣ⁻¹, x ϵR; x = ?
³√(8ˣ + 4) = ³√(2³ˣ + 4) = √4³ˣ⁻¹ = (√4³ˣ)/2, 2[³√(2³ˣ + 4)] = √4³ˣ = √2⁶ˣ
Let: y = 2³ˣ; 2[³√(y + 4)] = √(y²) = y, y³ = 8(y + 4), y³ - 8y - 32 = 0; y > 0
y³ - 64 - 8y + 32 = (y³ - 4³) - (8y - 32) = (y - 4)(y²+ 4y + 16) - 8(y - 4) = 0
(y - 4)(y²+ 4y + 16 - 8) = (y - 4)(y²+ 4y + 8) = 0, y = 2³ˣ, x ϵR; y²+ 4y + 8 > 0
y - 4 = 0, y = 4 = 2³ˣ, 2³ˣ = 4 = 2², 3x = 2; x = 2/3
Answer check:
³√(8²⸍3 + 4) = ³√8 = 2, √4³ˣ⁻¹ = √4³ˣ⁻¹ = √4 = 2; Confirmed
Final answer:
x = 2/3

[@RajeshKumar-wu7ox]

X=2/3 rest solution find out for dificult.

[@RealQinnMalloryu4]

{16^8+12 }= 28^8 7^4^2^3 7^12^2^2^3^1 1^11^12^1^1 2^1 (x ➖ 2x+1){ 64^27 ➖ 1} =64^26 8^8^2^13 2^32^3^2^13^1 1^11^12^1^1 2^1 (x ➖ 2x+1)

[@Quest3669]

X=2/3 only real soln
Rest is complex

[@yakupbuyankara5903]

2/3

[@abcekkdo3749]

X=2/3

[@kassuskassus6263]

One real solution x=2/3.

[@mulla_modi]

X=2/3