How to solve for x? : 82^2x+42^x+1=1+2^x

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Answer 1 · 2018-04-27T13:21:05

$x$ $x$ cannot be solved.

$8 \cdot 2^{2 x} + 4 \cdot 2^{x} + 1 = 1 + 2^{x}$ $8*2^(2x)+4*2^x+1=1+2^x$

Subtract $1$ $1$ from both sides,

$8 \cdot 2^{2 x} + 4 \cdot 2^{x} = 2^{x}$ $8*2^(2x)+4*2^x=2^x$

Subtract $2^{x}$ $2^x$ from both sides,

$8 \cdot 2^{2 x} + 3 \cdot 2^{x} = 0$ $8*2^(2x)+3*2^x=0$

Apply the power of a power law, where ${(a^{n})}^{m} = a^{n m}$ $(a^n)^m=a^(nm)$ ,

$8 \cdot {(2^{x})}^{2} + 3 \cdot 2^{x} = 0$ $8*(2^x)^2+3*2^x=0$

Let $2^{x}$ $2^x$ be $u$ $u$ ,

$8 u^{2} + 3 u = 0$ $8u^2+3u=0$

Factor,

$u (8 u + 3) = 0$ $u(8u+3)=0$

Solve,

$u = 0 or - \frac{3}{8}$ $u=0 or -3/8$

When $u = 0$ $u=0$ ,

$2^{x} = 0$ $2^x=0$ ( No solution as $2^{x}$ $2^x$ cannot be $0$ $0$ )

When $u = - \frac{3}{8}$ $u=-3/8$ ,

$2^{x} = - \frac{3}{8}$ $2^x=-3/8$ ( No solution as $2^{x}$ $2^x$ cannot be negative )

Hence, $x$ $x$ cannot be solved.

How to solve for x? : 8*2^2x+4*2^x+1=1+2^x