Question

Por favor alguém pode me ajudar ​

Answer 1

Explicação passo-a-passo:

1)

Lembre-se que:

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

• $\sf log_{b}~a^m=m\cdot log_{b}~a$

a)

$\sf log~30=log~(2\cdot3\cdot5)$

$\sf log~30=log~2+log~3+log~5$

$\sf log~30=0,30+0,47+0,69$

$\sf \red{log~30=1,46}$

b)

$\sf log~125=log~5^3$

$\sf log~125=3\cdot log~5$

$\sf log~125=3\cdot0,69$

$\sf \red{log~125=2,07}$

c)

$\sf log~100=log~(2^2\cdot5^2)$

$\sf log~100=log~2^2+log~5^2$

$\sf log~100=2\cdot log~2+2\cdot log~5$

$\sf log~100=2\cdot0,30+2\cdot0,69$

$\sf log~100=0,60+1,38$

$\sf \red{log~100=1,98}$

d)

$\sf log~81=log~3^4$

$\sf log~81=4\cdot log~3$

$\sf log~81=4\cdot0,47$

$\sf \red{log~81=1,88}$

e)

$\sf log~24=log~(2^3\cdot3)$

$\sf log~24=log~2^3+log~3$

$\sf log~24=3\cdot log~2+log~3$

$\sf log~24=3\cdot0,30+0,47$

$\sf log~24=0,90+0,47$

$\sf \red{log~24=1,37}$

f)

$\sf log~18=log~(2\cdot3^2)$

$\sf log~18=log~2+log~3^2$

$\sf log~18=log~2+2\cdot log~3$

$\sf log~18=0,30+2\cdot0,47$

$\sf log~18=0,30+0,94$

$\sf \red{log~18=1,24}$

g)

$\sf log~8=log~2^3$

$\sf log~8=3\cdot log~2$

$\sf log~8=3\cdot0,30$

$\sf \red{log~8=0,90}$

h)

$\sf log~10=log~(2\cdot5)$

$\sf log~10=log~2+log~5$

$\sf log~10=0,30+0,69$

$\sf \red{log~10=0,99}$

i)

$\sf log~16=log~2^4$

$\sf log~16=4\cdot log~2$

$\sf log~16=4\cdot0,30$

$\sf \red{log~16=1,20}$

j)

$\sf log~12=log~(2^2\cdot3)$

$\sf log~12=log~2^2+log~3$

$\sf log~12=2\cdot log~2+log~3$

$\sf log~12=2\cdot0,30+0,47$

$\sf log~12=0,60+0,47$

$\sf \red{log~12=1,07}$

2)

$\sf log_{2}~(-x^2+32)=4$

$\sf -x^2+32=2^4$

$\sf -x^2+32=16$

$\sf -x^2=16-32$

$\sf -x^2=-16~~~~\cdot(-1)$

$\sf x^2=16$

$\sf x=\pm\sqrt{16}$

• $\sf \red{x'=4}$

• $\sf \red{x"=-4}$

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