Question

$(2^x)^x=16 </span> <span>[tex] (10^x)^x^-^1=\frac{1}{10^6}$
$(4^x)^x= 256$
$2^x^2^-^7^x^+^1^2=1$
$3\cdot2^x^+^3=192$

Answer 1

 
 


a)  2^x^2 = 2^4 ==> x^2 = 4 ==> x = +/- 2

b) (10)^x^2 +1 = 10^-6  ==> x^2 = -6-1 ==>x^2=-7 ==>x=+/-V-7  sem solucão

c) 4^x^2 = 4^3 ==> x^2=3==>x=+/-V3

d) 2^x^2-7x+12 = 2^0

        x^2 -7x+12=0

delta= 49-12=1

x=7+/-V1 ==>x=7+/-1
       2.1                2

x1= 4  ; x2 = 3 

e) 3.2^x+3 = 192
        2^x+3 = 192/3
        2^x+3 = 64
        2^x+3 = 2^6
             x+3 =6
               x = 6-3 
               x = 3