Question

me ajudem pfv, é pra amanhã ​

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