Solve for x:
(I) 2^x + 4^x = A
Let A be any constant.
Note that 4=2^2. Then 4^x = (2^2)^x = (2^x)^2. Therefore:
(II) 2^x + (2^x)^2 = A
Let u = 2^x.
(III) u + u^2 = A
(IV) u^2 + u - A = 0
Solving the polynomial yields:
(V) u = (-1 ± √(1 + 4*A))/2
With u = 2^x:
(VI) 2^x = (-1 ± √(1 + 4*A))/2
Taking the logarithm of both sides, and with ln(B^C) = C ln B:
(VII) x * ln 2 = ln [ (-1 ± √(1 + 4*A))/2 ]
With ln(B/C) = ln B - ln C
(VIII) x * ln 2 = ln (-1 ± √(1 + 4*A) - ln 2
Solving for x:
(IX) x = [ ln (-1 ± √(1 + 4*A) ] / [ln 2] - 1
A General technique to solving A^x + B^x = C: (assuming B > A)
Let B = A^n. Then:
A^x + (A^x)^n = C
Let u = A^x, then solve the polynomial for u:
u + u^n - C = 0
Once the roots of u are found, then x = ln u/ln n
Solving this equation analytically is best when A, B, and C work "nicely".
Take care everyone, hope this helps and as always appreciate all the feedback!
This blog is property of Edward Shore. 2013
Wednesday, July 17, 2013
Solving 2^x + 4^x = A
Solve for x:
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