A Nice Exponential Equation | An Algebra Challenge

[@mohammedsaysrashid3587]

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

[@Quest3669]

X=2/3 only real soln Rest is complex

[@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.

[@yakupbuyankara5903]

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.

[@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

[@abcekkdo3749]

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

[@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)

[@kassuskassus6263]

One real solution x=2/3.

[@mulla_modi]

X=2/3