Solve the equation!:) 4^x + 2^(x+2) = 8
A alexs0uthern New member Joined Oct 22, 2014 Messages 2 Oct 22, 2014 #1 Solve the equation! 4^x + 2^(x+2) = 8
H HallsofIvy Elite Member Joined Jan 27, 2012 Messages 7,763 Oct 22, 2014 #2 \(\displaystyle 4= 2^2\) so \(\displaystyle 4^x= (2^2)^x= 2^{2x}= (2^x)^2\). \(\displaystyle 2^{x+ 2}= (2^x)(2^2)= 4(2^x)\) So your equation is the same as \(\displaystyle (2^x)^2+ 4(2^x)= 8\). If you let \(\displaystyle y= 2^x\), that is \(\displaystyle y^2+ 4y= 8\). Solve that quadratic equation for y, perhaps by "completing the square", then solve for x. The answers are NOT simple integer numbers!
\(\displaystyle 4= 2^2\) so \(\displaystyle 4^x= (2^2)^x= 2^{2x}= (2^x)^2\). \(\displaystyle 2^{x+ 2}= (2^x)(2^2)= 4(2^x)\) So your equation is the same as \(\displaystyle (2^x)^2+ 4(2^x)= 8\). If you let \(\displaystyle y= 2^x\), that is \(\displaystyle y^2+ 4y= 8\). Solve that quadratic equation for y, perhaps by "completing the square", then solve for x. The answers are NOT simple integer numbers!
