Question

25) Considerando log 2 -0,3, log 3 - 0,48 e log 5 = 0,7, calcule o valor de

a) log3 2
c) log2 5
b) log5 3
d) log3 100
e) log4 18
f) log36 0,5

ps; preciso de ajuda...​

Answer 1

Vamos utilizar as propriedades de logaritmo e a fatoração para reescrever os logaritmos de modo que possamos utilizar as informações dadas no enunciado para calcular o valor dos logaritmos.

a)

$Utilizando~a~propriedade~da~\underline{troca~de~base}\\\\\\\log_{_3}2~=~\dfrac{\log2}{\log3}\\\\\\\log_{_3}2~=~\dfrac{0,30}{0,48}\\\\\\\log_{_3}2~=~\dfrac{30}{48}\\\\\\\boxed{\log_{_3}2~=~\dfrac{5}{8}~~ou~~0,625}$

b)

$Utilizando~a~propriedade~da~\underline{troca~de~base}\\\\\\\log_{_2}5~=~\dfrac{\log5}{\log2}\\\\\\\log_{_2}5~=~\dfrac{0,70}{0,30}\\\\\\\log_{_2}5~=~\dfrac{70}{30}\\\\\\\boxed{\log_{_2}5~=~\dfrac{7}{3}~~ou~~2,333...}$

c)

$Utilizando~a~propriedade~da~\underline{troca~de~base}\\\\\\\log_{_5}3~=~\dfrac{\log3}{\log5}\\\\\\\log_{_5}3~=~\dfrac{0,48}{0,70}\\\\\\\log_{_5}3~=~\dfrac{48}{70}\\\\\\\boxed{\log_{_5}3~=~\dfrac{24}{35}}$

d)

$Utilizando~a~propriedade~da~\underline{troca~de~base}\\\\\\\log_{_3}100~=~\dfrac{\log100}{\log3}\\\\\\Fatorando~o~~ logaritmando~~''100''\\\\\\\log_{_3}100~=~\dfrac{\log\,(2\cdot2\cdot5\cdot5)}{\log3}\\\\\\Aplicando~a~propriedade~do~\underline{logaritmo~do~produto}\\\\\\\log_{_3}100~=~\dfrac{\log2~+~\log2~+~\log5~+~\log5}{\log3}\\\\\\\log_{_3}100~=~\dfrac{0,30~+~0,30~+~0,70~+~0,70}{0,48}\\\\\\\log_{_3}100~=~\dfrac{2,00}{0,48}$

$\log_{_3}100~=~\dfrac{200}{48}\\\\\\\boxed{\log_{_3}100~=~\dfrac{25}{6}~~ou~~4,166...}$

e)

$Utilizando~a~propriedade~da~\underline{troca~de~base}\\\\\\\log_{_4}18~=~\dfrac{\log18}{\log4}\\\\\\Fatorando~os~~ logaritmandos~~''18''~e~''4''\\\\\\\log_{_4}18~=~\dfrac{\log\,(2\cdot3\cdot3)}{\log\,(2\cdot2)}\\\\\\Aplicando~a~propriedade~do~\underline{logaritmo~do~produto}\\\\\\\log_{_4}18~=~\dfrac{\log2~+~\log3~+~\log3}{\log2~+~\log2}\\\\\\\log_{_4}18~=~\dfrac{0,30~+~0,48~+~0,48}{0,30~+~0,30}\\\\\\\log_{_4}18~=~\dfrac{1,26}{0,60}$

$\log_{_4}18~=~\dfrac{126}{60}\\\\\\\boxed{\log_{_4}18~=~\dfrac{21}{10}~~ou~~2,1}$

f)

$Utilizando~a~propriedade~da~\underline{troca~de~base}\\\\\\\log_{_{36}}0,5~=~\dfrac{\log0,5}{\log36}\\\\\\\log_{_{36}}0,5~=~\dfrac{\log\frac{1}{2}}{\log36}\\\\\\\log_{_{36}}0,5~=~\dfrac{\log2^{-1}}{\log36}\\\\\\Fatorando~o~~ logaritmando~~''36''\\\\\\\log_{_{36}}0,5~=~\dfrac{\log\,2^{-1}}{\log\,(2\cdot2\cdot3\cdot3)}\\\\\\Aplicando~a~propriedade~do~\underline{logaritmo~da~potencia}$

$\log_{_{36}}0,5~=~\dfrac{-1\cdot\log\,2}{\log\,(2\cdot2\cdot3\cdot3)}\\\\\\Aplicando~a~propriedade~do~\underline{logaritmo~do~produto}\\\\\\\log_{_{36}}0,5~=~\dfrac{-1\cdot\log2}{\log2~+~\log2~+~\log3~+~\log3}\\\\\\\log_{_{36}}0,5~=~\dfrac{-1\cdot0,30}{0,30~+~0,30~+~0,48~+~0,48}\\\\\\\log_{_{36}}0,5~=~\dfrac{-0,30}{1,56}\\\\\\\log_{_{36}}0,5~=\,-\dfrac{30}{156}\\\\\\\boxed{\log_{_{36}}0,5~=\,-\dfrac{5}{26}}$

Answer 2

Com base no estudo sobre logaritmos temos como resposta

• a)0,625;
• b)2,33;
• c)0,686;
• d)4,166;
• e)2,1;
• f)0,1923

Logaritmos

Dados dois números reais positivos, a e b diferentes de zero, algebricamente pode-se dizer que o logaritmo de b na base "a" é o expoente a que se tem de elevar "a" para que o resultado seja b.

• $\log _a\left(b\right)=c\:\Longleftrightarrow a^c=b$

O logaritmo é a operação inversa da exponencial. Quando os logaritmos são na base 10, chamam-se logaritmos decimais. No logaritmo decimal não se escreve a base.

Propriedades

1. Logaritmo de um produto:$\log _a\left(b\cdot c\right)=\log _a\left(b\right)+\log _a\left(c\right)$
2. Logaritmo de um quociente:$\log _a\left(\frac{b}{c}\right)=\log _a\left(b\right)-\log _a\left(c\right)$
3. Logaritmo de uma potência:$\log _a\left(b^n\right)=n\cdot \log _a\left(b\right)$
4. Logaritmo da unidade:$\log _a\left(1\right)=0$
5. Mudança de base em logaritmo:$\log _a\left(b\right)=\frac{\log _c\left(b\right)}{\log _c\left(a\right)}$

a) Vamos transformar $\log _3\left(2\right)$ para base 10 e para isso precisamos usar a última propriedade

• $\log _3\left(2\right)=\frac{\log _{10}\left(2\right)}{\log _{10}\left(3\right)}=\frac{0,3}{0,48}=0,625$

b)Vamos usar a mesma ideia da letra a)

• $\log _2\left(5\right)=\frac{\log _{10}\left(5\right)}{\log _{10}\left(2\right)}=\frac{0,7}{0,3}\approx 2,33$

c)

• $\log _5\left(3\right)=\frac{\log _{10}\left(3\right)}{\log _{10}\left(5\right)}=\frac{0,48}{0,7}\approx \:0,686$

d)Temos que 100 = 10², sendo assim

• $\log _3\left(100\right)\:=\:\log _3\left(10^2\right)\:=\:2.\log _3\left(10\right).$

• $\log _3\left(100\right)\:=\:\:2.\log _3\left(5\cdot 2\right)=2\left(\log _3\left(5\right)+\log _3\left(2\right)\right)$

• $\log _3\left(5\right)=\frac{\log _{10}\left(5\right)}{\log _{10}\left(3\right)}=\frac{0,7}{0,48}\approx \:1,458$

• $\log _3\left(100\right)=2\left(1,458\:+\:0,625\right)=4,166$

e)

• $\log _4\left(18\right)=\frac{\log _{10}\left(18\right)}{\log _{10}\left(4\right)}=\frac{\log _{10}\left(3^2\cdot 2\right)}{\log _{10}\left(2^2\right)}=$

• $=\frac{2\left(\log _{10}\left(3\right)\right)+\log _{10}\left(2\right)}{2\cdot \log _{10}\left(2\right)}=\frac{2\cdot 0,48\:+\:0,3}{2\cdot 0,3}=$

• $=\frac{0.30\:+\:0.96}{0,6}=2,1$

f)

• $\log _{36}\left(0,5\right)=\frac{\log _{10}\left(\frac{5}{10}\right)}{\log _{10}\left(36\right)}=\frac{\log _{10}\left(5\right)-\log _{10}\left(10\right)}{\log _{10}\left(36\right)}=$

• $=\log _{10}\left(5\right)-\log _{10}\left(10\right)=0,7-1=-0,3$

• $\log _{10}\left(36\right)=\log _{10}\left(2^2\cdot 3^2\right)=\log _{10}\left(2^2\right)+\log _{10}\left(3^2\right)=$

• $=2\cdot \log _{10}\left(2\right)+2\cdot \log _{10}\left(3\right)=0.60\:+\:0.96\:=\:1.56$

• $\log _{36}\left(0,5\right)=-\frac{0,3}{1,56}=\:0,1923$

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