Question

Qual o resultado desta equação exponencial (2^x)^x+4=32?

Answer 1

$(2^x)^{x+4}=32\to \\\\2^{x*(x+4)}=2^5\to x*(x+4)=5\to\\\\ x^{2} +4x=5\to \\\\ x^{2} +4x-5=0\\\\ a=1;b=4;c=-5\\\\\\ \Delta=b^2-4ac\to \Delta=4^2-4*1*(-5)\to \Delta=16+20\to \Delta=36\\\\ x' \neq x''\\\\ x= \frac{-b\pm \sqrt{\Delta} }{2a} \to x= \frac{-4\pm \sqrt{36} }{2*1} \to x= \frac{-4\pm 6 }{2} \to \\\\\\ x'= \frac{-4+ 6 }{2} \to x'= \frac{2 }{2} \to x'=1\\\\\\ x''= \frac{-4-6 }{2} \to x''= \frac{-10 }{2} \to x''=-5\\\\\\\\ S= (-5;1)$

Answer 2

Resposta:

x = 1

Explicação passo-a-passo:

2^(x+4)* x^(x+4) = 32  

16*2^x *x^(x+4) = 32  

2^x* x^(x+4) = 2  

2^1 * 1^(1+4) = 2*1^5 = 2*1 = 2  

x = 1