Evaluate. Textbook Answer: 16 Please show me in steps on how to solve this equation
K K_Swiss New member Joined Feb 8, 2008 Messages 28 Apr 8, 2008 #1 Evaluate. Textbook Answer: 16 Please show me in steps on how to solve this equation
pka Elite Member Joined Jan 29, 2005 Messages 11,995 Apr 8, 2008 #2 Re: Finding the Limit That given answer is incorrect. The limit is 1. (2x)(2x)=22x⇒22x−2x=2x(2x−1)\displaystyle \left( {2^x } \right)\left( {2^x } \right) = 2^{2x} \quad \Rightarrow \quad 2^{2x} - 2^x = 2^x (2^x - 1)(2x)(2x)=22x⇒22x−2x=2x(2x−1).
Re: Finding the Limit That given answer is incorrect. The limit is 1. (2x)(2x)=22x⇒22x−2x=2x(2x−1)\displaystyle \left( {2^x } \right)\left( {2^x } \right) = 2^{2x} \quad \Rightarrow \quad 2^{2x} - 2^x = 2^x (2^x - 1)(2x)(2x)=22x⇒22x−2x=2x(2x−1).
