How do you solve $4^{x} - 2^{x + 1} = 3$ $4^x-2^(x+1)=3$ ?

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How do you solve $4^{x} - 2^{x + 1} = 3$ $4^x-2^(x+1)=3$ ?

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Answer 1 · 2016-04-27T04:45:21

$x = \frac{log 3}{log 2}$ $x = log 3/log 2$ ..

Explanation:

Use $a^{m n} = {(a^{m})}^{n} = {(a^{n})}^{m} and a^{m + n} = a^{m} a^{n}$ $a^(mn)=(a^m)^n=(a^n)^m and a^(m+n)=a^ma^n$

$4^{x} = {(2^{2})}^{x} = 2^{2 x} = {(2^{x})}^{2}$ $4^x=(2^2)^x=2^(2x)=(2^x)^2$

The given equation is $u^{2} - 2 u - 3 = 0$ $u^2-2u-3=0$ , where $u = 2^{x}$ $u = 2^x$
The roots are u = $2^{x} = 3 and - 1$ $2^x =3 and -1$ . As $2^{x} > 0$ $2^x>0$ , for all $x, - 1$ $x, -1$ is inadmissible..

Now solve $2^{x} = 3$ $2^x=3$ , by equating the logarithms.

$x = \frac{log 3}{log 2}$ $x=log 3/log 2$ .

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How do you solve $4^{x} - 2^{x + 1} = 3$ $4^x-2^(x+1)=3$ ?

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How do you solve $4^{x} - 2^{x + 1} = 3$ $4^x-2^(x+1)=3$ ?