Question

moço inteligente me ajuda ​

Answer 1

01.

log[1/2] (1/32)=x

2^(-5) =2^(-x)

x=5

log[y] 81 =4

3^4 = y^4

y=3

log (xy/45) / log 3

[log x + log y -log 45]/ log 3

[log x + log y -log (3²*5)]/ log 3

[log x + log y -log 3² - log 5]/ log 3

[log 5 + log 3 -2*log 3 - log 5]/ log 3

[ -*log 3]/ log 3

=-1

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02.

a*b*c=1

log[b] c * log[c] a * log[a] b =1

logc/log b * loga /log c * log b /log a =1

=log a * log b * log c /(log a * log b * log c )

= log b * log c /( log b * log c )

= log c /( log c )

= 1 ==> CQP

# cqc ==>como queríamos provar

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03.

log[x] a =6 ==> log a=6 * log x

log[x] b=4 ==> log b= 4 * log x

log[x] c =2 ==> log c = 2 * log x

log[c] (abc)^(1/2)

=(1/2) * [log a +log b+log c]

=(1/2) * [6 * log x +4 * log x+2 * log x]

=(1/2) * [12 * log x]

= 6 * log x

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04.

log[a] a³* b²=m

log a³* b² / log a =m

(log a³ + log b²)=m * log a

(3*log a + log b²)=m * log a

log a *(m-3) =2* log b

log a/log b = 2/(m-3)

log[b] = 2/(m-3) é a resposta

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05.

log (a^(-1))^(1/2) - log b²

log a^(-1/2) - 2 log b

(1/2)* log a - 2*log b

=(1/2) * 2 -2 * 3

=1 -6

= -5

Answer 2

Resposta:

Explicação passo a passo:

$Quest\tilde{a} o\:1)$

$log_{\:y} \:81=4$

 $y^{4}=81$

 $y=\sqrt[4]{81}$

 $y=\sqrt[4]{3^{4} }$

 $y=3$      

$log_{\:\frac{1}{2} } \:(\frac{1}{32}) =x$

 $(\frac{1}{2}) ^{x} =\frac{1}{32}$

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

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

     $x=5$

$\frac{x\:.\:y}{45}=\frac{5\:.\:3}{45} =\frac{15}{45} =\frac{1}{3}$

    $Portanto...$

 $log_{\:3} \:(\frac{x\:.\:y}{45})=log_{\:3} \:(\frac{1}{3})=log_{\:3} \:(3^{-1}) =(-1)\:.\:log_{\:3} \:3=(-1)\:.\:1=-1$

$Quest\tilde{a} o\:2)$

$a\times\:b\times\:c=\:?$

$Vamos\:passar\:todos\:para\:a\:base\;10\:=>(base\;invisivel)$

$\frac{log\:c}{log\:b} \times\frac{log\:a}{log\:c}\times\:\frac{log\:b}{log\:a}=1$

$Por\:que\:deu\:1?$

$Resposta:\;Cancelando\:os\:termos\;iguais\:em\;cima\:e\:embaixo\:ok!$

$Quest\tilde{a} o\:3)$

$log_{\:x} \:a=6=>a=x^{6}$

$log_{\:x} \:b=4=>b=x^{4}$

$log_{\:x} \:c=2=>c=x^{2}$

$a.b.c=x^{6}.x^{4}.x^{2} =x^{12}$

$\sqrt{abc}=(abc)^{\frac{1}{2} }$

$log_{\:c} \:\sqrt{abc} =log_{\:c} \:(abc)^{\frac{1}{2} } =\frac{1}{2}\:.\:log_{\:c} \:(abc) =\frac{1}{2}\:.\:log_{\:c} \:(x^{12}) =\frac{1}{2}\:.\:12\:.\:log_{\:c}\:x=\\\\6\:.\:\frac{1}{log_{\:x}\:c } =6\:.\:\frac{1}{2}=3$

$Quest\tilde{a} o\:4)$

$log_{\:b}\:a=?$

$log_{\:a} \:(a^{3}\:.\:b^{2} )=m$

$log_{\:a} \:(a^{3}\:.\:b^{2} )=log_{\:a} \:a^{3} +log_{\:a}\:b^{2} =3\:.\:log_{\:a}\:a+2\:.\:log_{\:a}\:b=3\:.\:1+2\:.\:log_{\:a}\:b$

$Portanto...$

$3\:.\:1+2\:.\:log_{\:a}\:b=m$

$3+2\:.\:log_{\:a}\:b=m$

$2\:.\:log_{\:a}\:b=m-3$

$Observacao...$

$log_{\:a}\:b =\frac{1}{log_{\:b}\:a }$

$2\:.\:\frac{1}{log_{\:b}\:a }=m-3$

$\frac{2}{log_{\:b}\:a }=m-3$

$log_{\:b}\:a\: \;.\:(m-3)=2$

$log_{\:b}\:a=\frac{2}{(m-3)}$

$Quest\tilde{a} o\:5)$

$log\:a=2$

$log\:b=3$

$log(\:\frac{\sqrt{\frac{1}{a} } }{b^{2} }) =$

$log\:\sqrt{\frac{1}{a} } \:\:-\:\:log\:b^{2} =$

$log\:(\frac{1}{a}) ^{\frac{1}{2} } \:-\:2\:.\:log\:b=$

$Observacao:$

$\frac{1}{a} =a^{-1}$

$\frac{1}{2}\:.\: log\:(a^{-1}) -2\:.\:log\:b=$

$\frac{1}{2}\:.\:(-1)\:.\:log\:a-2\:.\:log\:b=$

$-\frac{1}{2}\:.\:log\:a\:-\:2\:.\:log\:b=$

$-\frac{1}{2}\:.\:(2)\:-\:2\:.\:(3)=$

    $-\frac{2}{2}\:-\:6=$

     $-1-6=$

         $-7$

          $Ufa!$

          $Bye!$

         $Jean318$