Question

dadas:log 2=0,301, log 3 = 0,477 e log 5 = 0,65 , calcule .

a) log 32=
b) log $\sqrt[4]{125}$ =
c) log $\frac{25}{6}$ =
d) log 114 =

Answer 1

a)
$\log32=log2^5=5log2=5(0,301)=1,505$

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b)
$log \sqrt[4]{125} =log \sqrt[4]{5^3} =log 5^{ \frac{3}{4} } = \frac{3}{4} log5= \\ \\ \frac{3}{4} (0,65)= \frac{1,95}{4} =0,4875$

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c)
$log \frac{25}{6} =log25-log6= \\ \\ log5^2-(log2+log3)=$

$2log5-(0,301+0,477)= \\ \\ 2(0,65)-(0,778)= \\ \\ 1,30-0,778=0,522$

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d)
$log144=log2^4.3^2=4log2+2log3= \\ \\ 4(0,301)+2(0,477)= \\ \\ 1,204+0,954=2,158$

Answer 2

Explicação passo-a-passo:

Lembre-se que:

• $\sf log_{b}~(a\cdot c)=log_{b}~a+log_{b}~c$

• $\sf log_{b}~(a\div c)=log_{b}~a-log_{b}~c$

• $\sf log_{b}~a^m=m\cdot log_{b}~a$

• $\sf \sqrt[c]{a^b}=a^{\frac{b}{c}}$

a)

$\sf log~32=log~2^5$

$\sf log~32=5\cdot log~2$

$\sf log~32=5\cdot0,301$

$\sf \red{log~32=1,505}$

b)

$\sf log~\sqrt[4]{125}=log~\sqrt[4]{5^3}$

$\sf log~\sqrt[4]{125}=log~5^{\frac{3}{4}}$

$\sf log~\sqrt[4]{125}=\dfrac{3}{4}\cdot log~5$

$\sf log~\sqrt[4]{125}=\dfrac{3}{4}\cdot0,65$

$\sf log~\sqrt[4]{125}=\dfrac{1,95}{4}$

$\sf \red{log~\sqrt[4]{125}=0,4875}$

c)

$\sf log~\Big(\dfrac{25}{6}\Big)=log~25-log~6$

$\sf log~\Big(\dfrac{25}{6}\Big)=log~5^2-log~(2\cdot3)$

$\sf log~\Big(\dfrac{25}{6}\Big)=2\cdot log~5-(log~2+log~3)$

$\sf log~\Big(\dfrac{25}{6}\Big)=2\cdot log~5-log~2-log~3$

$\sf log~\Big(\dfrac{25}{6}\Big)=2\cdot0,65-0,301-0,477$

$\sf log~\Big(\dfrac{25}{6}\Big)=1,3-0,301-0,477$

$\sf \red{log~\Big(\dfrac{25}{6}\Big)=0,522}$

d)

$\sf log~144=log~(2^4\cdot3^2)$

$\sf log~144=log~2^4+log~3^2$

$\sf log~144=4\cdot log~2+2\cdot log~3$

$\sf log~144=4\cdot0,301+2\cdot0,477$

$\sf log~144=1,204+0,954$

$\sf \red{log~144=2,158}$