(5 + (24)^1/2)^x + (5 - (24)^1/2)^x = 10 , solve for x

Question

24^1/2 means 24 to the power 1/2

Answer 1

already answered at your previous question.. link is https://www.wyzant.com/resources/answers/218427/5_24_1_2_x_5_24_1_2_x_10_solve_for_x

Answer 2

(5 + (24)^1/2)^x + (5 - (24)^1/2)^x = 10................ (1) Given

Let's assume, 5 + (24)^1/2=a and 5 - (24)^1/2=b which gives, a.b=5²-(24^1/2)²=25-24=1

substituting a and b in (1),

a^x+b^x=10 ...............(2)

=>a^x+1b+ab^x+1=10 .... multiplying both sides by ab

=>a^x+2b+ab^x+2=10 .... multiplying both sides by ab

Also dividing both sides in (2) by ab

=>a^x-1b+ab^x-1=10 .... dividing both sides by ab
=>a^x-2b+ab^x-2=10   
Since such multiplication and division cannot change the constant value of the right hand side of the equation, values of a and b are very small which, in turn, indicates that x is very small.

Again expanding for even values of x,

a^x=5^x+x.5^x-1.24^1/2+[x(x-1)/2!]5^x-2.24+[x(x-1)(x-3)/3!]5^x-3.24^3/2+.....+24^x/2

b^x=5^x -x.5^x-1.24^1/2+[x(x-1)/2!]5^x-2.24-[x(x-1)(x-3)/3!]5^x-3.24^3/2+.....+24^x/2

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Adding,

a^x+b^x=2[5^x+[x(x-1)/2!]5^x-2.24+.....+24^x/2]

=>2[5^x+[x(x-1)/2!]5^x-2.24+.....+24^x/2]=10 ..................from (2)

=>5^x+[x(x-1)/2!]5^x-2.24+.....+24^x/2=5 ..................dividing both sides by 2

=>5^x-1+[x(x-1)/2!]5^x-3.24+.....+24^x/2/5=1 ..................dividing both sides by 5

=>x is very small, hence the terms that are multiples of x or having x as its power are negligible, and 

5^x-1≈1

=>5^x-1≈5⁰ ..... comparing w.r.t equal bases

=>x-1≈0

=>x≈1.... adding 1 to both sides... hence solved.