## Wednesday, July 17, 2013

### Solving 2^x + 4^x = A

Solve for x:

(I) 2^x + 4^x = A

Let A be any constant.

Then:

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!

Eddie

This blog is property of Edward Shore. 2013

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