Question

urgentee. me ajudem pfv​

Answer 1

Explicação passo-a-passo:

a)

$\sf log_{4}~x=2$

$\sf x=4^2$

$\sf x=4\cdot4$

$\sf \red{x=16}$

b)

$\sf log_{x}~81=4$

$\sf x^4=81$

$\sf x=\sqrt[4]{81}$

$\sf \red{x=3}$

c)

$\sf log_{x}~64=2$

$\sf x^2=64$

$\sf x=\sqrt[2]{64}$

$\sf \red{x=8}$

d)

$\sf log_{x}~8=3$

$\sf x^3=8$

$\sf x=\sqrt[3]{8}$

$\sf \red{x=2}$

e)

$\sf log_{x}~\dfrac{1}{16}=2$

$\sf x^2=\dfrac{1}{16}$

$\sf x=\sqrt[2]{\dfrac{1}{16}}$

$\sf \red{x=\dfrac{1}{4}}$

f)

$\sf log_{2}~x=5$

$\sf x=2^5$

$\sf x=2\cdot2\cdot2\cdot2\cdot2$

$\sf \red{x=32}$

g)

$\sf log_{\frac{1}{3}}~27=x$

$\sf \Big(\dfrac{1}{3}\Big)^x=27$

$\sf (3^{-1})^x=3^3$

$\sf 3^{-x}=3^3$

Igualando os expoentes:

$\sf -x=3~~~\cdot(-1)$

$\sf \red{x=-3}$

h)

$\sf log_{x}~32=-5$

$\sf x^{-5}=32$

$\sf (x^{-5})^{-1}=32^{-1}$

$\sf x^5=\dfrac{1}{32}$

$\sf x=\sqrt[5]{\dfrac{1}{32}}$

$\sf \red{x=\dfrac{1}{2}}$

Answer 2

Explicação passo-a-passo:

Sabendo que:

$log_{a}(x) = b \\ \ \ {a}^{b} = x$

Vamos lá:

a)

$log_{4}(x) = 2 \\ \ \ {4}^{2} = x \\ x = 16$

b)

$log_{x}(81) = 4 \\ {x}^{4} = 81 \\ {x}^{4} = {3}^{4} \\ x = 3$

c)

$log_{x}(64) = 2 \\ \ \ {x}^{2} = 64 \\ {x}^{2} = {8}^{2} \\ x = 8$

d)

$log_{x}(8) = 3 \\ \ \ {x}^{3} = 8 \\ {x}^{3} = {2}^{3} \\ x = 2$

e)

$log_{x}( \frac{1}{16} ) =2 \\ {x}^{2} = \frac{1}{16} \\ {x}^{2} = \frac{1}{4} ^{2} \\ x = \frac{1}{4}$

f)

$log_{2}(x) = 5 \\ \ \ {2}^{5} = x \\ x = 32$

g)

$log_{ \frac{1}{3} }(27) = x \\ \ \ { \frac{1}{3} }^{x} = 27 \\ { \frac{1}{3} }^{x} = { \frac{1}{27} }^{ - 1} \\ \frac{1}{3} ^{x} = { \frac{1}{3} }^{ - 3} \\ x = - 3$

h)

$log_{x}(32) = - 5 \\ \ \ {x}^{ - 5} = 32 \\ {x}^{ - 5} = { \frac{1}{2} }^{ - 5} \\ x = \frac{1}{2}$

Bons estudos^_^