A Fascinating Exponential Equation Challenge | You Should Try!

[@RyanLewis-Johnson-wq6xs]

Input 49/16^2 + 49/32^2 + 49/64^2 = 1/2^2 + 1/32^2 + 1/64^2 Result True Left hand side 49/16^2 + 49/32^2 + 49/64^2 = 1029/4096 Right hand side 1/2^2 + 1/32^2 + 1/64^2 = 1029/4096

[@AmirgabYT2185]

x=2

[@michaeldoerr5810]

The answe is x= 2. Allow me to make an observation: at 7:50 mark, instead of aubteacting deom the LHS, you could just cancel out the a^2+a+1 with knowledge that it is irreducible over R, and subtract 49 to a^3-a^2+1. After that it is just reducibg that polynomial to difference between a cube and a square. That is just my thought.

[@RyanLewis-Johnson-wq6xs]

49/16^2+49/32^2+49/64^2=1/2^2+1/32^2+1/64^2 x=2 final answer

[@Quest3669]

X= 2

[@ManojkantSamal]

(49/2^4x)+(1/32^x)+(48/32^x)+(1/64^x)+(48/64^x)=(1/2^x)+(1/32^x)+(1/64^x) So, (49/2^4x)+(48/2^5x)+(48/2^6x)=(1/2^x) So (48/2^4x)+(48/2^5x)+(48/2^6x)=(1/2^x)-(1/2^4x) So 48{(1/2^4x)+(1/2^5x)+(1/2^6x)}={(2^3x)-1}/(2^4x) 48[{(2^2x)+(2^x)+1}/(2^6x)]={(2^3x)-1}/(2^4x)

[@RashmiRay-c1y]

Let a=1/2^x. Then 48a^6+48a^5+49a^4-a=0. a is not zero. So, 48a^5+48a^4+49a^3-1=0 > (4a-1)(a^2+a+1)(12a^2+3a+1)=0. The only real solution is a=1/4 > x=2.

[@michaeldoerr5810]

Just one question did you use synthetic division??? I have noticed that this simple procedure has been posted on two or three @infyGyan problems before going live. Do you also use Wolfram???

[@RealQinnMalloryu4]

{49x+49x ➖ }/{16x+16x ➖ }= 96x^2/32x^3+:{49x+49x ➖ }/{32x+32x ➖ }+{49x+49x ➖}/{64x+64x ➖ } {96x,^2/32x^2+96x^2/64x^2}= {192x^4/96x^4 +96x^2/128x^2}=288x^6/234x^6 = 1.54x^1 6^9 3^2^3^3 1^2^3^1 2^3 (x ➖ 3x+2) .{1x+1x ➖ }/{2x+2x ➖ }=2x^2/4x^2 {1x+1x ➖}/{32x+32x ➖ }={2x^2x^2+2x^2/64x^2}=4x^4/68x^4 {1x+1x ➖}/{64x+64x ➖}= {4x^4/68x^4+2x^2/128x^2 }= 6x^6/196x^6 3^2x^3^2/10^10^8^12x^3^2 1^1x1^1/2^5^2^5^8^3^4x^1^1 /1^1^1^1^2^3^3^2^2 /1^1^1^3^1^2 3^2 (x ➖ 3x+2)

[@kassuskassus6263]

x=2