Question

Não consigo por um anexo, me responde por favor !!

Answer 1

$d) \\ 16^{x} = \sqrt{4}==\ \textgreater \ (4^2)^{x} =4 ^\frac{1}{2} ==\ \textgreater \ 2x= \frac{1}{2} ==\ \textgreater \ x=\frac{1}{4} \\ e) \\ \frac{1}{9} ^x= \sqrt[4]{27} ==\ \textgreater \ ( \frac{1}{3^2})^x= \sqrt[4]{3^3} \\ 3^{-2x} =3^{ \frac{3}{4} }==\ \textgreater \ -2x= \frac{3}{4} ==\ \textgreater \ x= \frac{-3}{8} \\ f) \\ (2 \sqrt{2})^x=16==\ \textgreater \ (2*2^{ \frac{1}{2} })^x=2^4==\ \textgreater \ 2^{ \frac{3x}{2} }=2^{4} \\ \frac{3x}{2} =4==\ \textgreater \ x=8/3$


$g) (\frac{1}{4}) ^x=2 \sqrt{2} ==\ \textgreater \ (\frac{1}{2^2}) ^x=2*2^{ \frac{1}{2}} \\ 2^{-2x}=2^{ \frac{^3}{2}} ==\ \textgreater \ -2x=\frac{^3}{2}==\ \textgreater \ x= \frac{-3}{4}$


$h) \\ 5^x=125^ \frac{^1}{2} ==\ \textgreater \ 5^x=(5^3)^ \frac{^1}{2}==\ \textgreater \ 5^x=5^ \frac{3}{2} \\ x= \frac{3}{2}$


$i) \\ 27^x= \sqrt{3} ==\ \textgreater \ 3^{3x}=3^{ \frac{1}{2} }==\ \textgreater \ 3x= \frac{1}{2} ==\ \textgreater \ x=\frac{1}{6}$



$j) \\ 7^x= \sqrt{49} ==\ \textgreater \ 7^{x}=49^{ \frac{1}{2} } \\ 7^{x}=7^{2* \frac{1}{2} }==\ \textgreater \ x= 1$