Question

1)usando a definição de logaritmo, calcule:
c)$log(10000)$
$d) log\binom{1}{2} 32$
$e) log_{10}(0.01)$
$f) log_{2}(0.5)$

$g) log_{2}( \sqrt{8} )$
$h) log_{4}( \sqrt{32} )$
$i) log_{ \binom{1}{4} }16$
​

Answer 1

Me acompanhe:

$c) \: log(10000) \\log(10000) = x \\ {10}^{4} = {10}^{x} \\ x = 4 \\ </p><p>d) log\binom{1}{2} 32 \\ log_{ \frac{1}{2} }(32) = x \\ { (\frac{1}{2} )}^{x} = {2}^{5} \\ {2}^{ - x} = {2}^{5} \\ x = - 5 \\ </p><p>e) log_{10}(0.01) \\ log(0.01) = x \\ {10}^{x} = {10}^{ - 3} \\ x = - 3 \\ </p><p>f) log_{2}(0.5) \\ log_{2}( \frac{1}{2} ) = x \\ {2}^{x} = \frac{1}{2} \\ {2}^{x} = {2}^{ - 1} \\ x = - 1 \\ </p><p>g) log_{2}( \sqrt{8} ) \\ log_{2}( \sqrt{8} ) = x \\ {2}^{x} = {8}^{ \frac{1}{2} } \\ {2}^{x} = {2}^{ \frac{3}{2} } \\ x = \frac{ 3 }{2} \\ h) log_{4}( \sqrt{32} ) \\ log_{4}( \sqrt{ 32} ) = x \\ {4}^{x} = {32}^{ \frac{1}{2} } \\ { {2}^{2} }^{x} = { {2}^{ \frac{5}{2} } }\\ x = \frac{5}{2} \\ i) log_{ \binom{1}{4} }16 \\ log_{ \frac{1}{4} }(16) = x \\ ( \frac{1}{2} ) {}^{x} = 16 \\ {2}^{ - x} = {2}^{4} \\ x = - 4$

Espero ter lhe ajudado! ❤❤☺