Let 5^x=t. Then, the equation becomes t^4-5t^3+6t^2-5t+1=0, which upon dividing both sides bt t^2, yields t^2+1/t^2 -5(t+1/t)+6 = 0 > (t+1/t)^2 -5(t+1/t) + 4 = 0 > t+1/t= 1,4. But t+1/t=1 does not lead to a real x. So, with t+1/4 = 4, t = 2 +/-√3 > x = [ln(2+/-√3)]/ln5.
Let 5^x=a and solve a^4-5a^3+6a^2-5a+1=0. Divide both sides by a^2. We get (a^2+1/a^2 -5(a+1/a)+6=0. Let's make another COV, u=a+1/a. Then, we have to solve u^2-5u+4=0. For u=1, the discriminant is negative, and for u=4, we'll get a=2+or-sqrt3. So, x=log(2+or-sqrt3)/log5.
Let 5^x=t. Then, the equation becomes t^4-5t^3+6t^2-5t+1=0, which upon dividing both sides bt t^2, yields t^2+1/t^2 -5(t+1/t)+6 = 0 > (t+1/t)^2 -5(t+1/t) + 4 = 0 > t+1/t= 1,4. But t+1/t=1 does not lead to a real x. So, with t+1/4 = 4, t = 2 +/-√3 > x = [ln(2+/-√3)]/ln5.
Let 5^x=t; t^4-5t^3+6t^2-5t+1=0;divide with t^2 then
y=t+1/t;y^2-5y+4=0;x=log(2±√3) /log5👍
Let 5^x=a and solve a^4-5a^3+6a^2-5a+1=0. Divide both sides by a^2. We get (a^2+1/a^2 -5(a+1/a)+6=0. Let's make another COV, u=a+1/a. Then, we have to solve u^2-5u+4=0. For u=1, the discriminant is negative, and for u=4, we'll get a=2+or-sqrt3. So, x=log(2+or-sqrt3)/log5.
X= log ( 2+ - √3)/ log 5
x=log (2+√3)/log 5 ,log(2-√3)/log5, another solution is comlex not valid.
5ˣ = a
(a² + 5)/(5a - 1) - 1/(a²) = 1
a²(a² + 5) - (5a - 1) = a²(5a - 1)
a²(a² + 5) - (a² + 1)(5a - 1) = 0
a²(a² + 5) + (a² + 1)(1 - 5a) = 0
a⁴ + 5a² + a² - 5a³ + 1 - 5a = 0
a⁴ - 5a³ + 6a² - 5a + 1 = 0
a² - 5a + 6 - 5/a + 1/a² = 0
(a² + 1/a²) - 5(a + 1/a) + 6 = 0
(a + 1/a)² - 5(a + 1/a) + 4 = 0
a + 1/a = u
u² - 5u + 4 = 0
(u - 4)(u - 1) = 0
u = 4
a + 1/a = 4
a² - 4a + 1 = 0
a = 2 ± √3
5ˣ= 2 ± √3
*x = log₅(2 ± √3)*
u = 1
a + 1/a = 1
a² - a + 1 = 0 => complex solution
x = log₅(2 ± √3)
verification
[(2 ± √3)² + 5]/[5(2 ± √3) - 1) - 1/(2 ± √3)² = 1
(12 ± 4√3)/(9 ± 5√3) - 1/(7 ± 4√3) = 1
(12 ± 4√3)(7 ± 4√3) - (9 ± 5√3) = (9 ± 5√3)(7 ± 4√3)
(12 ± 4√3)(7 ± 4√3) = (8 ± 4√3)(9 ± 5√3)
(3 ± √3)(7 ± 4√3) = (2 ± √3)(9 ± 5√3)
33 ± 19√3 = 33 ± 19√3
10x+{5+5 ➖ }/5{x+x ➖ }{1+1 ➖}(1)^2 = {10x+10 } /5{x^2+2} ➖ 1= 20/5^2x^2 ➖ 1= 20x/{10x^2 ➖ 1}= 20x/9x^2 =2.2x^2 1.1x^2 1x^2 (x ➖ 2x+1). {1+1 ➖}/10x =2/10x 2^1/2^5x 1^1/2^1x 2^1x (x ➖ 2x+1) .
χ=[log(2+ _ριζα3)]/log5