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

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Answer 1 · 2016-04-27T04:45:21

$x = log 3/log 2$ ..

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

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

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

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

$x=log 3/log 2$ .

How do you solve 4^x-2^(x+1)=34^x-2^(x+1)=3?