Respuesta :

[tex]\bf \textit{Logarithm Cancellation Rules} \\\\ log_a a^x = x\qquad \qquad a^{log_a x}=x \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \log_6(\sqrt[3]{6})\implies \log_6(6^{\frac{1}{3}})\implies \cfrac{1}{3} \\\\[-0.35em] ~\dotfill\\\\ \log_2(64)\implies \log_2(2^6)\implies 6 \\\\[-0.35em] ~\dotfill\\\\ -3\log_5(25)\implies -3\log_5(5^2)\implies -3(2)\implies -6 \\\\[-0.35em] ~\dotfill[/tex]

[tex]\bf \log_2(\sqrt[4]{8})\implies \log_2(\sqrt[4]{2^3})\implies \log_2(2^{\frac{3}{4}})\implies \cfrac{3}{4} \\\\[-0.35em] ~\dotfill\\\\ \log_3\left( \frac{1}{81} \right)\implies \log_3\left( \frac{1}{3^4} \right)\implies \log_3(3^{-4})\implies -4[/tex]

Answer:

3/4 goes with [tex]\log_2(8^\frac{1}{4})[/tex]

-4 goes with [tex]\log_3(\frac{1}{81})[/tex]

-6 goes with [tex]-3\log_5(25)[/tex]

1/3 goes with [tex]\log_6(6^\frac{1}{3})[/tex]

Step-by-step explanation:

[tex]\log_6(6^\frac{1}{3})=\frac{1}{3}\log_6(6)=\frac{1}{3}\cdot 1=\frac{1}{3}[/tex]

[tex]\log_2(64)=6 \text{ since } 2^6=64[/tex]

[tex]-3\log_5(25)=-3(2)=-6 \text{ since } 5^2=25[/tex]

[tex]\log_2(8^\frac{1}{4})=\frac{1}{4}\log_2(8)=\frac{1}{4}\log_2(2^3)=\frac{1}{4}\cdot (3)\log_2(2)=\frac{1}{4} \cdot 3 \cdot 1=\frac{3}{4} [/tex]

[tex] \log_3(\frac{1}{81})=\log_3(\frac{1}{3^4})=\log_3(3^{-4})=-4\log_3(3)=-4(1)=-4[/tex]

Here are few rules I used:

[tex]\log_a(b)=x \text{ means } a^x=b[/tex]

[tex]\log_a(a)=1 [/tex]

[tex]\log_a(b^r)=r \log_a(b)[/tex]

Otras preguntas