Question

me ajudemm por favor ​

Answer 1

Resposta e explicação passo a passo:

1.a)

     $5^x^+^3=1\\\\5^x^+^3=5^0\\\\x+3=0\\\\x=-3$

b)

   $2^{x^2}^-^9=1\\\\2^{x^2}^-^9=2^0\\\\x^2-9=0\\\\x^2=9\\\\x=^+_-9$

c)

   $7^3^x^+^7=7\\\\7^3^x^+^7=7^1\\\\3x+7=1\\\\3x=1-7\\\\3x=-6\\\\x=\frac{-6}{3}\\\\x=-2$

d)

   $3^{x^2}^{-3}=3\\\\3^{x^2}^{-3}=3^1\\\\x^{2}-3=1\\\\x^2=1+3\\\\x=^+_-\sqrt{4}\\\\x= ^+_-2$

e)

     $\frac{1}{9^x} =27\\\\9^{-x}=27\\\\3^{-3x}=3^3\\\\-3x=3\\\\x=\frac{3}{-3}\\\\x=-1$

f)

   $\frac{1}{25^x}=\sqrt{125}\\\\25^{-x}=5^{\frac{3}{2}}\\\\5^{-2x}=5^{\frac{3}{2}}\\\\-2x=\frac{3}{2}\\\\x=-2.\frac{3}{2}\\\\x=\frac{-6}{2}\\\\x=-3$

2. a)

   $2.3^{x-2}=162\\\\2.3^{x-2}=3^4.2\\\\x-2=4\\\\x=6$

b)

   $3.5^{x-1}=75\\\\3.5^{x-1}=5^2.3\\\\x-1=2\\\\x=3$

c)

   $10.2^{x+3}=10\\\\10.2^{x+3}=10.1\\\\2^{x+3}=2^0\\\\x+3=0\\\\x=-3$

3. a)

$10^{x^2-2x-2}=10\\\\x^2-2x-2=1\\\\x^2-2x-3=0\left \{ {{x'=3} \atop {x''=-1}} \right.$

b)