What is the answer to 4^x+4^x+2^x=1 ?

The answer to 4^x+4^x+2^x=1 is x=-1

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4^x+4^x+2^x=1

4^{x} + 4^{x} + 2^{x} = 1

x = - 1

Solution steps

4^{x} + 4^{x} + 2^{x} = 1

Apply exponent rules

(2^{x})^{2} + (2^{x})^{2} + 2^{x} = 1

Rewrite the equation with

2^{x} = u

(u)^{2} + (u)^{2} + u = 1

Solve

u^{2} + u^{2} + u = 1 : u = \frac{1}{2}, u = - 1

u = \frac{1}{2}, u = - 1

Substitute back

u = 2^{x},

solve for

x

Solve

2^{x} = \frac{1}{2} : x = - 1

Solve

2^{x} = - 1 :

No Solution for

x \in R

x = - 1

3^{x + 4} = 9^{x + 13}

x, 8^{2 x} \cdot 3^{- \frac{1}{2} \cdot y} = 36

\frac{243}{3125} = (\frac{3}{5})^{3 n - 1}

2 + 5 e^{2 x - 5} = 3

8^{x} - 5^{x} = 0