1. $\log _a(m n)=\log _a m+\log _a n \quad a>0, a \neq 1 \quad$&$m>0, n>0$

2. $\log _a\left(\frac{m}{n}\right)=\log _a m-\log _a n \quad a>0, a \neq 1$ & $m>0, n>0$

3. $\log _a m=\frac{\log _c m}{\log _c a} \quad a>0, a \neq 1, c>0, c \neq 1$ & $m > 0$

4. $\log _a m^n=n \log _a m, a>0, a \neq 1, \quad n \in \mathbb{R}$ & $m>0$

Solve for $x$

$\log _a(x+3)+\log _a(x-3)=\log _a 16$

Soln. $\log _a (x+3)(x-3)=\log _a 16$

(Multiplication law)

$\log _a m=\log _a n \Leftrightarrow m=n$

$(x+3)(x-3)=16$

$x^2-9=16$

$x^2-25=0$

$(x-5)(x+5)=0$

$x=-5, 5$

$(x+3)(x-3)$

case I. $x=-5 (-5+3)(-5-3) =(-2)(-8)=16$

$\log _a(-5+3)+\log _a(-5-3)=$

$\log _a(-2)+\log _a(-8)$ $x=-5$ is not solving the expression.

Case 2. $x=5$

$\log _a(x+3)+\log _a(x-3)=\log _a 16$

$\log _a 8+\log _a 2=\log _a 16$

$\log _a(8 \times 2)=\log _a 16,$

$\log_a 16 = \log_a 16$

$x=5$ is the solution far given equation.

Solve: $2 \log _{10} x=1+\log _{10}\left(x+\frac{11}{10}\right)$

Soln. $1=\log _{10} 10$

$\log _{10} x^2=\log _{10} 10+\log _{10}\left(x+\frac{11}{10}\right)$ $\times \log _{10} x^2$

$=\log _{10}\left(10\left(x+\frac{11}{10}\right)\right)$

$x^2=10\left(x+\frac{11}{10}\right)$

$\log _a m=\log _a n \Leftrightarrow m=n$

$x^2=10 x+11$

$x^2-10 x-11=0$

$x^2-11 x+x-11=0$

$(x-11)(x+1)=0$

$x=-1,11$

$2 \log _{10}(-1)^{-}$is not defined

Case 1. $\quad x=-1$ is not solution to this equation.

Case 2. $\underline{x=11} \quad 2 \log _{10} 11=1+\log _{10}\left(11+\frac{11}{10}\right)$

$\log _{10} 121 =1+\log _{10}\left(\frac{110+11}{10}\right)$

$=1+\log _{10}\left(\frac{121}{10}\right)$

$=1+\log _{10}(121)-\log _{10} 10$

$x=11$ is the only solution to this equation:

$\log_a \left(\frac{x+y}{6}\right)=\frac{1}{2}(\log_a x+\log_a y)$

Show that $\frac{x}{y}+\frac{y}{x}=34$

Solution. $\log_a (x+y) \stackrel{?}{=} \log_a x+\log_a y$

$\log_a (x y)=\log_a x+\log_a y$

$\log (ab) = \log (cd)$

$\log _a\left(\frac{x+y}{6}\right)=\frac{1}{2} \log _a(x y) \rightarrow 2 \log _a\left(\frac{(x+y)}{6}\right)=\log _a(x y)$

$\log_a\left(\frac{(x+y)^2}{36}\right)= \log_a x y$

$\frac{(x+y)^2}{36} =xy, \quad \rightarrow x^2+y^2+2 x y=36 x y,$

$x^2+y^2=34 x y \quad \rightarrow \frac{x^2+y^2}{x y}=34,$

$\frac{x}{y}+\frac{y}{x} = 34 ,$

$\log_am = \log_an, \Leftrightarrow m=n$

4. If $a^2-12 a b+4 b^2=0$.

Prove that $\quad \log (a+2 b)=$

$\frac{1}{2}(\log a+\log b)+2 \log 2$

Solution. Given$\quad a^2-12 a b+4 b^2=0$

To show. $\log (a+2 b)=\frac{1}{2}(\log a+\log b)+2 \log 2$

$\frac{1}{2}$ Power law, $(\log a+\log b)$ Multiplication law, 2 $\log 2$ power law

$\log ()=\log (\quad)$

$\log (a+2 b) =\log (a b)^{1 / 2}+\log 2^2$

$\log (a+2 b) =\log (4 \sqrt{a b})$

$(a+2 b) =4 \sqrt{a b} \quad \rightarrow (a+2 b)^2 =16 a b$

$a^2+4 b^2+4 a b =16 a b \Rightarrow a^2+4 b^2-12 a b =0$

5. Show that $\frac{2}{3}<\log _{10} 5<\frac{3}{4}$.

Soln. $\frac{2}{3}<\log _{10} 5$

$\frac{2}{3} \times 1<\log _{10} 5$

$\frac{2}{3} \log _{10} 10<\log _{10} 5$

$2 \log _{10} 10<3 \log _{10} 5$

$\log _{10} 10^2<\log _{10} 5^3$

$\log _{10} 100<\log _{10} 125$

$m<n \Leftrightarrow \log _a m<\log _a n$

$\log _{10} 5<\frac{3}{4}$

$4 \log _{10} 5<3 \log _{10} 10 \quad \rightarrow \log _{10} 5^4<\log _{10} 10^3$

$\log _{10} 625<\log _{10} 1000 \Rightarrow 625<1000$

Show that

$\log _a\left(\frac{50}{147}\right)$

$=\log _a 2+2 \log _a 5-\log _a 3-2 \log _a 7$

Soln. RHS $=\log 2+2 \log 5-\log 3-2 \log 7$

$=\log 2+\log 5^2-\log 3-\log 7^2$

$=\log \left(\frac{2 \times 5^2}{3 \times 7^2}\right)$

$=\log \left(\frac{2 \times 25}{3 \times 49}\right)=\log \left(\frac{50}{147}\right)$

Solve for $x, \log _4(x-1)=\log _2(x-3)$

Soln. $2^2=4,4^{1 / 2}=2$

$\log _a b=\frac{\log _c b}{\log _c a}$

RHS $=\log _2(x-3)=\log _{4^{1 / 2}}(x-3)$

$=\frac{\log _4(x-3)}{\log _4 4^{1 / 2}}$ COB $=\frac{\log _4(x-3)}{1 / 2 \log _4 4} =2 \log _4(x-3)$

$\log _4(x-1) =\log _4(x-3)^2$

$\Leftrightarrow \quad(x-1) =(x-3)^2$

$x-1 =x^2-6 x+9$

$x^2-7 x +10=0$

$\Leftrightarrow(x-5)(x-2)=0$

$x=5,2$ are the possible solutions

Case 1.$x=2 \quad \log (x-3)=\log \sqrt{4} (-1)$

$x=5 \quad \log _2(x-3)=\log _2 2=1$

$x=5$