Question

alguem me ajuda
preciso dos calculos

Answer 1

a)

$2^x = 8 \\ \\ 2^x = 2^3 \\ \\ x = 3$

b)

$3^{x+5} = 3^5 \\ \\ x+5 = 5 \\ \\ x = 0$

c)

$243 = 3^{x+2} \\ \\ 3^5 = 3^{x+2} \\ \\ 5 = x+2 \\ \\ x = 3$

d)

$2^{5x} = 128 \\ \\ 2^{5x} = 2^7 \\ \\ 5x = 7 \\ \\ x = \frac{7}{5}$

e)

$5^x = \frac{1}{125} \\ \\ 5^x = 125^{-1} \\ \\ 5^x = (5^3)^{-1} \\ \\ 5^x = 5^{-3} \\ \\ x = -3$

f)

$7^{x+1} = 7 \\ \\ x + 1 = 1 \\ \\ x = 0$

g)

$4^x = \sqrt{2} \\ \\ 4^x = 2^{ \frac{1}{2} } \\ \\ 2^{2x} = 2^{ \frac{1}{2} } \\ \\ 2x = \frac{1}{2} \\ \\ x = \frac{ \frac{1}{2} }{2} \\ \\ x = \frac{1}{4}$

h)

$9^x = \sqrt[3]{3} \\ \\ 3^{3x} = 3^{ \frac{1}{3} } \\ \\ 3x = \frac{1}{3} \\ \\ x = \frac{ \frac{1}{3} }{3} \\ \\ x = \frac{1}{9}$

i)

$8^x = \sqrt[5]{2^4} \\ \\ 2^{3x} = 2^{\frac{4}{5}} \\ \\ 3x = \frac{4}{5} \\ \\ x = \frac{ \frac{4}{5} }{3} \\ \\ x = \frac{4}{15}$

j)

$\frac{1}{9^x} = \sqrt{3} \\ \\ 9^{-x} = 3^{ \frac{1}{2} } \\ \\ 3^{-2x} = 3^{ \frac{1}{2} } \\ \\ -2x = \frac{1}{2} \\ \\ x = \frac{ \frac{1}{2} }{-2} \\ \\ x = - \frac{1}{4}$

l)

$\sqrt[5]{4^x} = \frac{1}{ \sqrt{2} } \\ \\ 4^{ \frac{x}{5} }= (2^{ \frac{1}{2} })^{-1} \\ \\ 2^{ \frac{2x}{5} } = 2^{- \frac{1}{2} } \\ \\ \frac{2x}{5} = -\frac{1}{2} \\ \\ 4x = -5 \\ \\ x = - \frac{5}{4}$

m)

$2^{3x+1} = 2 \\ \\ 3x+1 = 1 \\ \\ 3x = 0 \\ \\ x = 0$

n)

$5^{ x^{2} -9 } = 1 \\ \\ 5^{ x^{2} -9 } = 5^0 \\ \\ x^{2} - 9 = 0 \\ \\ x^{2} = 9 \\ \\ x' = 3 \\ x'' = -3$

o)

$2^{ x^{2} -x} = 1 \\ \\ 2^{ x^{2} -x} = 2^0 \\ \\ x^{2} -x = 0 \\ \\ x(x - 1) = 0 \\ \\ x' = 0 \\ x'' = 1$

p)

$7^{ x^{2} } = 7 \\ \\ x^{2} = 1 \\ \\ x' =1 \\ x'' = -1$