$\log_2{4}+\log_2{8}=\quad$ $\log_2{12}\qquad$ $\log_2{16}\qquad$ $\log_2{32}\qquad$ $\log_4{8}\qquad$ $\log_8{4}$
$\log_3{1}+\log_3{27}=\quad$ $\log_3{28}\qquad$ $\log_3{27}\qquad$ $\log_3{81}\qquad$ $\log_1{27}\qquad$ $\log_27{3}$
$\log_5{\frac{1}{5}}+\log_5{25}=\quad$ $\log_5{5}\qquad$ $\log_5{25}\qquad$ $\log_5{25\frac{1}{5}}\qquad$ $\log_{25}{\frac{1}{5}}\qquad$ $\log_{25}{5}$
$\log_4{2}+\log_4{8}=\quad$ $\log_4{\frac{1}{4}}\qquad$ $\log_4{2}\qquad$ $\log_4{8}\qquad$ $\log_4{10}\qquad$ $\log_4{16}$
The examples follow the rule,
$\log_a{x}+\log_a{y}=\quad$ $\log_a{\left(x+y\right)}\qquad$
$\log_a{\left(x-y\right)}\qquad$
$\log_a{\left(xy\right)}\qquad$
$\log_a{\left(x^y\right)}$
Let's investigate why $\log_2{4}+\log_2{8}=\log_2{\left(32\right)}$.
Let $4=2^m$. Then $m=$
Let $8=2^n$. Then $n=$
$\log_2{\left(32\right)}=\log_2{\left(4\times8\right)}=\log_2{\left(2^2\times2^3\right)}=\log_2{\bigl(2}$$2+$ $\bigl)=2+3=\log_2{4}+\log_2{8}$
Let's investigate why $\log_a{x}+\log_a{y}=\log_a{\left(xy\right)}$.
$\log_a{\left(x \times y\right)}=\log_a{\Bigl(a^{\log_a{}}}$ $\times a$ $\Bigl)=\log_a{\Bigl(a}$ $\Bigl)=\log_a$ $+\log_a$
$\log_2{16}-\log_2{8}=\quad$ $\log_2{1}\qquad$ $\log_2{2}\qquad$ $\log_2{4}\qquad$ $\log_2{8}\qquad$ $\log_2{16}$
$\log_3{81}-\log_3{27}=\quad$ $\log_3{1}\qquad$ $\log_3{\frac{4}{3}}\qquad$ $\log_3{3}\qquad$ $\log_3{9}\qquad$ $\log_3{54}$
$\log_5{5}-\log_5{25}=\quad$ $\log_5{\left(-20\right)}\qquad$ $\log_5{\frac{1}{5}}\qquad$ $\log_5{\frac{1}{25}}\qquad$ $\log_5{\frac{1}{125}}\qquad$ $\log_5{5}$
$\log_4{2}-\log_4{16}=\quad$ $-\log_4{\left(8\right)}\qquad$ $\log_4{\left(-14\right)}\qquad$ $\log_4{\frac{1}{4}}\qquad$ $\log_4{\frac{1}{8}}\qquad$ $\log_4{\frac{1}{16}}$
The examples follow the rule,
$\log_a{x}-\log_a{y}=\quad$ $\log_a{\left(x+y\right)}\qquad$
$\log_a{\left(x-y\right)}\qquad$
$\log_a{\left(xy\right)}\qquad$
$\log_a{\left(\frac{x}{y}\right)}$
Let's investigate why $\log_3{81}-\log_3{27}=\log_3{3}$.
Let $81=3^m$. Then $m=$
Let $27=3^n$. Then $n=$
$\log_3{\left(3\right)}=\log_3{\left(\frac{81}{27}\right)}=\log_3{\left(\frac{3^4}{3^3}\right)}=\log_3{\Bigl(3}$$4-$ $\Bigl)=4-3=\log_3{81}-\log_3{27}$
Let's investigate why $\log_a{x}-\log_a{y}=\log_a{\left(\frac{x}{y}\right)}$.
$\log_a{\left(\frac{x}{y}\right)}=\log_a{\Biggl(}$ $a$ / $a$ $\Biggl)=\log_a{a}$ $=\log_a$ $-\log_a$