What is the answer to solvefor f,f(2^x+1)=8^x+3*4^x+2^x-5 ?

The answer to solvefor f,f(2^x+1)=8^x+3*4^x+2^x-5 is f=(8^x+3*4^x+2^x-5)/(2^x+1)

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solvefor f,f(2^x+1)=8^x+3*4^x+2^x-5

solve for

f, f (2^{x} + 1) = 8^{x} + 3 \cdot 4^{x} + 2^{x} - 5

f = \frac{8 ^{x} + 3 \cdot 4 ^{x} + 2 ^{x} - 5}{2 ^{x} + 1}

Solution steps

f (2^{x} + 1) = 8^{x} + 3 \cdot 4^{x} + 2^{x} - 5

Divide both sides by

2^{x} + 1

f = \frac{8 ^{x} + 3 \cdot 4 ^{x} + 2 ^{x} - 5}{2 ^{x} + 1}

a^{2} x - a = x + 1

x, f (x + 7) = 2 x - 3

\frac{( 5 x )}{3} - \frac{( x - 2 )}{4} = \frac{21}{4} - (x - \frac{( 2 x - 1 )}{3})

x, 8 x + 7 y = 876 x - 28 y - 234

(x + 1) (x + 2) (x + 5) - x^{2} (x + 8) = 27

What is the answer to solvefor f,f(2^x+1)=8^x+3*4^x+2^x-5 ?
The answer to solvefor f,f(2^x+1)=8^x+3*4^x+2^x-5 is f=(8^x+3*4^x+2^x-5)/(2^x+1)