Question

considerando que log2=0,30, log 3= 0,48 e log5=0,69 resolva as questão.
a)log 32
b)log3 5
c)log5 2
d)log9 45
e)log2 18sobre27

Answer 1

a) log32 = log(2^5) = 5log2 = 5*0,3 = 1,5

b) $log_35 = \frac{log5}{log3} = \frac{0,69}{0,48} = 1,43$

c) $log_52 = \frac{log2}{log5} = \frac{0,3}{0,69} = 0,43$

d) $log_945 = \frac{log45}{log9} = \frac{log(5*3^2)}{log3^2} = \frac{log5+2log3}{2log3} = \frac{0,69+2*0,48}{2*0,48} = \frac{1,65}{0,96} = 1,71$

e) $log_2 \frac{18}{27} = log_218 - log_227 =\\ \frac{log18}{log2} - \frac{log27}{log2} \\ \frac{log18-log27}{log2} \\ \frac{log(3^2*2)-log(3^3)}{0,3} \\ \frac{2log3+log2-3log3}{0,3} \\ \frac{2*0,48 + 0,3 -3*0,48}{0,3} \\ \frac{0,96+0,3-1,44}{0,3} \\ \frac{-0,18}{0,3} = -0,6$

Answer 2

Olá Deco,


$\mathsf{\ell og(2)\approx0,3}\\\mathsf{\ell og(3)\approx 0,48}\\\mathsf{\ell og(5)\approx0,69}\\\\\mathsf{A-}\\\mathsf{\ell og(32)\Rightarrow \ell og(2^5)\Rightarrow 5\cdot(0,3)=\boxed{\mathsf{1,5}}}\\\\\mathsf{B-}\\\mathsf{\ell og_3(5)\Rightarrow\dfrac{\ell og(5)}{\ell og(3)}\Rightarrow\dfrac{0,69}{0,48}=\boxed{\mathsf{1,4375}}}\\\\\mathsf{C-}\\\mathsf{\ell og_5(2)\Rightarrow\dfrac{\ell og(2)}{\ell og(5)}\Rightarrow\dfrac{0,3}{0,69}\approx\boxed{\mathsf{0,434}}}$

$\mathsf{D-}\\\mathsf{\ell og_9(45)\Rightarrow \dfrac{\ell og(3\cdot3\cdot5)}{\ell og(3\cdot3)}\Rightarrow\dfrac{0,48+0,48+0,69}{0,48+0,48}\Rightarrow\dfrac{1,65}{0,96}=\boxed{\mathsf{1,71875}}}}}\\\\\mathsf{E-}\\\mathsf{\ell og_2\Big(\dfrac{18}{27}\Big)\Rightarrow\ell og_2\Big(\dfrac{2\cdot\diagup\!\!\!\!3\cdot\diagup\!\!\!\!3}{3\cdot\diagup\!\!\!\!3\cdot\diagup\!\!\!\!3}\Big)\Rightarrow\dfrac{\ell og\Big(\dfrac{2}{3}\Big)}{\ell og(2)}\Rightarrow\dfrac{0,3-0,48}{0,3}\Rightarrow\dfrac{-0,18}{0,3}}\\\\=$

$\boxed{\mathsf{-0,6}}$

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