mathematics-class-12-unit-05-chapter-02-logarithm-

### <Success style="font-size:25px"> Logarithm L-2  </Success>
### <primary style="font-size:24px"> Laws of logarithms </primary>
<hr>
<main>

- $ \log _a a=1$
- $\left(\because \log _a a=x \Rightarrow a^x=a^1 \Rightarrow x=1\right)$

- $ \log _{1 / a} a=-1$ 
- $\left(\because \log _{1 / a} a=x \Rightarrow\left(\frac{1}{a}\right)^x=a^1 \Rightarrow x=-1\right)$

- $ \log _{a} 1 =0 $
- $\left(\because \log _\alpha 1=x \Rightarrow a^x=1=a^0 \Rightarrow x=0\right)$ 
- $ a^{\log _a b}=b$

</div>
<div style='float: right; width:1%;'>

</div>
</main>
<hr>
<footer style = "font-size: 18px"> $\rightarrow$ Logarithm lecture - 2 $\rightarrow$ <span style = "color:blue"> Laws of logarithms </span> $\rightarrow$ Laws of logarithms $\rightarrow$ Proofs </footer>

### <Success style="font-size:25px"> Logarithm L-2  </Success>
### <primary style="font-size:24px"> Laws of logarithms </primary>
<hr>
<main>

- Assume $a>0, \quad a \neq 1, \quad m>0, n>0, \quad c>0, c \neq 1$

- 1. $\log _a(m n)=\log _a m+\log _a n \quad$ (Multiplication law)
 
- 2. $\log _a(m / n)=\log _a m-\log _a n \quad$ (Quotient law)

- 3. $\log _a m=\frac{\log _c m}{\log _c a} \quad$ (Change of base)

- 4. $\log _a\left(m^n\right)=n \log _a m$ (Power law):

</div>
</main>
<hr>
<footer style = "font-size: 18px">Logarithm lecture - 2 $\rightarrow$ Laws of logarithms $\rightarrow$ <span style = "color:blue"> Laws of logarithms </span> $\rightarrow$ Proofs $\rightarrow$ Proofs </footer>

### <Success style="font-size:25px"> Logarithm L-2  </Success>
### <primary style="font-size:24px"> Proofs </primary>
<hr>
<main>

- <ins>**1. Multiplication law**</ins>

- $ \log _a(m n)  =\log _a m+\log _a n$

- $ \log _a m  =x(\text { say }) $

- $ a^x  =m $,$ \log _a n=y \text { (say) }$

- ${a y}=n$   ,$ m n  =a^x a^y $ ,$ m n  =a^{x+y} $

- $ \log _a m n=x+y$,$ \log _a(m n)=\log _a m+\log _a n$,$ \log _a(12)=\log _a(4 \times 3)=\log _a 4+\log _a 3$ .

<ins>Remark</ins>

- $ \log _a(m n p q)=\log _a m+\log _a n+\log _a p+\log _a q$

- $ \log _a({m n}{p q})=\log _a({m n})+\log _a(p q)$

</div>
<div style='float: right; width:1%;'>

</div>
</main>
<hr>
<footer style = "font-size: 18px">Laws of logarithms $\rightarrow$ Laws of logarithms $\rightarrow$ <span style = "color:blue"> Proofs </span> $\rightarrow$ Proofs $\rightarrow$ Proofs </footer>

### <Success style="font-size:25px"> Logarithm L-2  </Success>
### <primary style="font-size:24px"> Proofs </primary>
<hr>
<main>

- <ins>**2. Quotient law**</ins>

- $ \log _a\left(\frac{m}{n}\right)=\log _a m-\log _a n \text {. }$,$ \log _a m=x(\text { say }) \quad $

- $ a^x=m$,$ \frac{m}{n}=\frac{a^x}{a^y}=a^{x-y}$

- $\log _a n=y(\text { say })$,$ a y=n $ ,$ \frac{a^x}{a^y}=a^x(\frac{1}{a^y})=a^x a^{-y}$,$ =a^{x - y}$

- $ \frac{m}{n}=a^{x-y}$

- $ \log _a\left(\frac{m}{n}\right)  =x-y $

- $ =\log _a m-\log _a n$

- $ \log _a\left(\frac{m}{n p q}\right)  =\log _a m-\log _a(n p q)$

- $ =\log _a m-\log _a n-\log _a p-\log _a q$

- $\log _a\left(\frac{3}{15 \times 4}\right)=\log _{a^3} 3-\log _a 15-\log _a 4$

</div>
</main>
<hr>
<footer style = "font-size: 18px">Laws of logarithms $\rightarrow$ Proofs $\rightarrow$ <span style = "color:blue"> Proofs </span> $\rightarrow$ Proofs $\rightarrow$ Proofs </footer>

### <Success style="font-size:25px"> Logarithm L-2  </Success>
### <primary style="font-size:24px"> Proofs </primary>
<hr>
<main>

- **3. Change of base.**
	
- $\log _a m=\frac{\log _c m}{\log _c a}$,$\left(\log _{\underline{a}} m\right)\left(\log _c a\right)=\log _c m$

- $ \log _a m=x(\text { say }) $ ,$\underline{a}^x=m$ ,$ \quad \log _c a=y(\text { say })$,$  \quad c^y=a, .$

- $ \quad \log _c m=z(\text { say })$
- $ \quad c^z=m $    
- $(c^y)^x = m = c^z$
- $c^{x y} =c^z$       
- $\Rightarrow x y  =z$,$ \quad\left(\log _a r=\log _a s \Leftrightarrow r=s\right)$

- $ \left(\log _a m\right)\left(\log _c a\right)=\log _c m $,$ \log _x a=\frac{\log _a a}{\log _a x}=\frac{1}{\log _a x}$

- $ \log _x a=\frac{1}{\log _a x}$

- $ a=1 \quad \log _x 1=\frac{1}{\log _1 x} \rightarrow \text { (Not defined) }$

</div>
<div style='float: right; width:1%;'>

</div>
</main>
<hr>
<footer style = "font-size: 18px">Proofs $\rightarrow$ Proofs $\rightarrow$ <span style = "color:blue"> Proofs </span> $\rightarrow$ Proofs $\rightarrow$ Problems on laws of logarithms </footer>

### <Success style="font-size:25px"> Logarithm L-2  </Success>
### <primary style="font-size:24px"> Proofs </primary>
<hr>
<main>

- <ins>**4. Power law**</ins>

- $ \log _a\left(m^n\right)=n \log _a m \text {. } $

- $ \log _a m =x(\text { say }) \quad$
- $ a^x=m \ldots (*)$

- $\log _{a}\left(m^n\right)=y(\text { say })$
- $ a^y=m^n \ldots (**)$

- from equation $(*)$ and $( * *)$

- $ a^y=\left(a^x\right)^n =a^{n x}$

- $ \therefore y=n x $

- $ \log _a\left(m^n\right)=n \log _a m$

- $ \log _2 4-\log _2\left(2^2\right)=2 \log _2 2=2 . [\log _a a=1]$

</div>
</main>
<hr>
<footer style = "font-size: 18px">Proofs $\rightarrow$ Proofs $\rightarrow$ <span style = "color:blue"> Proofs </span> $\rightarrow$ Problems on laws of logarithms $\rightarrow$ Problems on laws of logarithms </footer>

### <Success style="font-size:25px"> Logarithm L-2  </Success>
### <primary style="font-size:24px"> Problems on laws of logarithms </primary>
<hr>
<main>

- <ins>**1. Simplify**</ins>

- $ \frac{1}{2} \log _5 36+2 \log _5 7-\frac{1}{2} \log _5 12$     
- $ =\log _5(36)^{1 / 2}+\log _5 7^2-\log _5(12)^{1 / 2} \quad\left(\log _a\left(m^n\right)=n \log _a m\right)$

- $ =\log _5\left(\frac{6 \times 49}{2 \sqrt{3}}\right)=\log _5(49 \sqrt{3}) \text {. }$       
- (Multiplication \& Quotient law)

</div>
</main>
<hr>
<footer style = "font-size: 18px">Proofs $\rightarrow$ Proofs $\rightarrow$ <span style = "color:blue"> Problems on laws of logarithms </span> $\rightarrow$ Problems on laws of logarithms $\rightarrow$ Problems on laws of logarithms </footer>

### <Success style="font-size:25px"> Logarithm L-2  </Success>
### <primary style="font-size:24px"> Problems on laws of logarithms </primary>
<hr>
<main>

- <ins>**2. Evaluate**</ins>

- $ \log _5\left(\frac{\sqrt[4]{25}}{625}\right) \quad  \sqrt[4]{25}=(25)^{1 / 4} \text {Power law}$

- $ =  \log _5\left((25)^{1 / 4}\right)-\log _5 625  \text { (Quotient law })$

- $ =  \frac{1}{4} \log _5 25-\log _5 625$

- $ =  \frac{1}{4} \log _5\left(5^2\right)-\log _5\left(5^4\right)$
- $ =  \frac{2}{4} \log _5 5-4 \log _5 5  $

- $ =  \left(\frac{2}{4}-4\right) \log _5 5  \quad \quad \left(\log _a a=1\right)$
- $ =  \left(\frac{2}{4}-4\right)$
- $ =\frac{-14}{4}$ 
- $ =\frac{-7}{2} $

</div>
</main>
<hr>
<footer style = "font-size: 18px">Proofs $\rightarrow$ Problems on laws of logarithms $\rightarrow$ <span style = "color:blue"> Problems on laws of logarithms </span> $\rightarrow$ Problems on laws of logarithms $\rightarrow$ Thank you </footer>

### <Success style="font-size:25px"> Logarithm L-2  </Success>
### <primary style="font-size:24px"> Problems on laws of logarithms </primary>
<hr>
<main>

- <ins>**3. Evaluate.**</ins>

- $ \log _{10} 2+16 \log _{10}\left(\frac{16}{15}\right)+12 \log _{10}\left(\frac{25}{24}\right)+7 \log _{10}\left(\frac{81}{80}\right)$

- $ =\log _{10} 2+\log _{10}\left(\frac{16}{15}\right)^{16}+\log _{10}\left(\frac{25}{24}\right)^{12}+\log _{10}\left(\frac{81}{80}\right)^7 \quad $
- $ \text { (Power law) }$
- $ \text { (Quotient law) }$
- $ \text { (Multiplication) }$

- $ =\log _{10}\left(2 \times\left(\frac{16}{15}\right)^{16} \times\left(\frac{25}{24}\right)^{12} \times\left(\frac{81}{80}\right)^7\right)$

- $ =\log _{10}\left(\frac{2 \times\left(2^4\right)^{16} \times\left(5^2\right)^{12} \times\left(3^4\right)^7}{3^{16} \times 5^{16} \times 3^{12} \times\left(2^3\right)^{12} \times\left(2^4\right)^7 \times 5^7}\right)$

- $ =\log _{10}\left(\frac{2 \times 2^{64} \times 5^{24} \times 3^{28}}{3^{28} \times 5^{23} \times 2^{64}}\right)$
- $ =\log _{10}(2 \times 5)=\log _{10} 10=1$ .

</div>
</main>
<hr>
<footer style = "font-size: 18px">Problems on laws of logarithms $\rightarrow$ Problems on laws of logarithms $\rightarrow$ <span style = "color:blue"> Problems on laws of logarithms </span> $\rightarrow$ Thank you $\rightarrow$  </footer>