Question

PLZ HELP
Pre-calculus

Answer 1

3/4 goes with
`\log_2(8^(1)/(4))`

-4 goes with
`\log_3((1)/(81))`

-6 goes with
`-3\log_5(25)`

1/3 goes with
`\log_6(6^(1)/(3))`

Explanation:

`\log_6(6^(1)/(3))=(1)/(3)\log_6(6)=(1)/(3)\cdot 1=(1)/(3)`

`\log_2(64)=6 \text{ since } 2^6=64`

`-3\log_5(25)=-3(2)=-6 \text{ since } 5^2=25`

`\log_2(8^(1)/(4))=(1)/(4)\log_2(8)=(1)/(4)\log_2(2^3)=(1)/(4)\cdot (3)\log_2(2)=(1)/(4) \cdot 3 \cdot 1=(3)/(4)`

`\log_3((1)/(81))=\log_3((1)/(3^4))=\log_3(3^(-4))=-4\log_3(3)=-4(1)=-4`

Here are few rules I used:

`\log_a(b)=x \text{ means } a^x=b`

`\log_a(a)=1`

`\log_a(b^r)=r \log_a(b)`

Answer 2

`\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^{(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`

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