mathematics-class-12-unit-05-chapter-03-logarithm-mccat6cmtqw

<h3 id="success-stylefont-size25px-logarithm-l-3-success-1"><Success style="font-size:25px"> Logarithm L-3 </Success></h3>
<h3 id="primary-stylefont-size24px-review-of-laws-of-logarithms-primary"><primary style="font-size:24px"> Review of Laws of logarithms </primary></h3>
<hr>
<main>
<div style='float: left; width:99%;font-size:30px; text-align:left;'>
<ul>
<li>
<ol>
<li>$\log _a(m n)=\log _a m+\log _a n \quad a>0, a \neq 1 \quad $&$ m>0, n>0$</li>
</ol>
</li>
<li>
<ol start="2">
<li>$\log _a\left(\frac{m}{n}\right)=\log _a m-\log _a n \quad a>0, a \neq 1 $ & $ m>0, n>0$</li>
</ol>
</li>
<li>
<ol start="3">
<li>$\log _a m=\frac{\log _c m}{\log _c a} \quad a>0, a \neq 1, c>0, c \neq 1 $ & $ m > 0$</li>
</ol>
</li>
<li>
<ol start="4">
<li>$\log _a m^n=n \log _a m,  a>0, a \neq 1, \quad n \in \mathbb{R}  $ & $ m>0$</li>
</ol>
</li>
</ul>
</div>
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</main>
<hr>
<footer style = "font-size: 18px"> $\rightarrow$ Logarithms lecture - 3 $\rightarrow$ <span style = "color:blue"> Review of Laws of logarithms </span> $\rightarrow$ Problem 1 $\rightarrow$ Problem 1 </footer>

<h3 id="success-stylefont-size25px-logarithm-l-3-success-3"><Success style="font-size:25px"> Logarithm L-3 </Success></h3>
<h3 id="primary-stylefont-size24px-problem-1-primary-1"><primary style="font-size:24px"> Problem 1 </primary></h3>
<hr>
<main>
<div style='float: left; width:99%;font-size:21px; text-align:left;'>
<ul>
<li>
<p>case I. $x=-5 (-5+3)(-5-3)  =(-2)(-8)=16$</p>
</li>
<li>
<p>$\log _a(-5+3)+\log _a(-5-3)=$</p>
</li>
<li>
<p>$\log _a(-2)+\log _a(-8)$ $x=-5$ is not solving the expression.</p>
</li>
<li>
<p>Case 2. $x=5$</p>
</li>
<li>
<p>$\log _a(x+3)+\log _a(x-3)=\log _a 16 $</p>
</li>
<li>
<p>$\log _a 8+\log _a 2=\log _a 16 $</p>
</li>
<li>
<p>$\log _a(8 \times 2)=\log _a 16, $</p>
</li>
<li>
<p>$\log_a 16 = \log_a 16$</p>
</li>
<li>
<p>$x=5$ is the solution far given equation.</p>
</li>
</ul>
</div>
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<hr>
<footer style = "font-size: 18px">Review of Laws of logarithms $\rightarrow$ Problem 1 $\rightarrow$ <span style = "color:blue"> Problem 1 </span> $\rightarrow$ Problem 2 $\rightarrow$ Problem 2 </footer>

<h3 id="success-stylefont-size25px-logarithm-l-3-success-4"><Success style="font-size:25px"> Logarithm L-3 </Success></h3>
<h3 id="primary-stylefont-size24px-problem-2-primary"><primary style="font-size:24px"> Problem 2 </primary></h3>
<hr>
<main>
<div style='float: left; width:99%;font-size:22px; text-align:left;'>
<ul>
<li>
<p>Solve: $ 2 \log _{10} x=1+\log _{10}\left(x+\frac{11}{10}\right)$</p>
</li>
<li>
<p>Soln.  $ 1=\log _{10} 10 $</p>
</li>
<li>
<p>$ \log _{10} x^2=\log _{10} 10+\log _{10}\left(x+\frac{11}{10}\right)$ $\times \log _{10} x^2$</p>
</li>
<li>
<p>$=\log _{10}\left(10\left(x+\frac{11}{10}\right)\right) $</p>
</li>
<li>
<p>$ x^2=10\left(x+\frac{11}{10}\right) $</p>
</li>
<li>
<p>$ \log _a m=\log _a n  \Leftrightarrow m=n $</p>
</li>
<li>
<p>$ x^2=10 x+11   $</p>
</li>
<li>
<p>$ x^2-10 x-11=0 $</p>
</li>
<li>
<p>$ x^2-11 x+x-11=0 $</p>
</li>
<li>
<p>$ (x-11)(x+1)=0$</p>
</li>
<li>
<p>$ x=-1,11 $</p>
</li>
</ul>
</div>
</main>
<hr>
<footer style = "font-size: 18px">Problem 1 $\rightarrow$ Problem 1 $\rightarrow$ <span style = "color:blue"> Problem 2 </span> $\rightarrow$ Problem 2 $\rightarrow$ Problem 3 </footer>

<h3 id="success-stylefont-size25px-logarithm-l-3-success-5"><Success style="font-size:25px"> Logarithm L-3 </Success></h3>
<h3 id="primary-stylefont-size24px-problem-2-primary-1"><primary style="font-size:24px"> Problem 2 </primary></h3>
<hr>
<main>
<div style='float: left; width:99%;font-size:22px; text-align:left;'>
<ul>
<li>
<p>$2 \log _{10}(-1)^{-}$is not defined</p>
</li>
<li>
<p>Case 1. $\quad x=-1$ is not solution to this equation.</p>
</li>
<li>
<p>Case 2. $\underline{x=11} \quad 2 \log _{10} 11=1+\log _{10}\left(11+\frac{11}{10}\right)$</p>
</li>
<li>
<p>$\log _{10} 121  =1+\log _{10}\left(\frac{110+11}{10}\right) $</p>
</li>
<li>
<p>$ =1+\log _{10}\left(\frac{121}{10}\right) $</p>
</li>
<li>
<p>$ =1+\log _{10}(121)-\log _{10} 10$</p>
</li>
<li>
<p>$x=11$ is the only solution to this equation:</p>
</li>
</ul>
</div>
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</div>
</main>
<hr>
<footer style = "font-size: 18px">Problem 1 $\rightarrow$ Problem 2 $\rightarrow$ <span style = "color:blue"> Problem 2 </span> $\rightarrow$ Problem 3 $\rightarrow$ Problem 4 </footer>

<h3 id="success-stylefont-size25px-logarithm-l-3-success-6"><Success style="font-size:25px"> Logarithm L-3 </Success></h3>
<h3 id="primary-stylefont-size24px-problem-3-primary"><primary style="font-size:24px"> Problem 3 </primary></h3>
<hr>
<main>
<div style='float: left; width:99%;font-size:25px; text-align:left;'>
<ul>
<li>
<p>$\log_a \left(\frac{x+y}{6}\right)=\frac{1}{2}(\log_a x+\log_a y)$</p>
</li>
<li>
<p>Show that $\frac{x}{y}+\frac{y}{x}=34$</p>
</li>
<li>
<p>Solution.  $ \log_a (x+y) \stackrel{?}{=} \log_a x+\log_a y $</p>
</li>
<li>
<p>$ \log_a (x y)=\log_a x+\log_a y$</p>
</li>
<li>
<p>$\log (ab) = \log (cd)$</p>
</li>
<li>
<p>$ \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)$</p>
</li>
<li>
<p>$\log_a\left(\frac{(x+y)^2}{36}\right)= \log_a x y$</p>
</li>
<li>
<p>$\frac{(x+y)^2}{36} =xy, \quad \rightarrow  x^2+y^2+2 x y=36 x y, $</p>
</li>
<li>
<p>$x^2+y^2=34 x y \quad \rightarrow \frac{x^2+y^2}{x y}=34, $</p>
</li>
<li>
<p>$ \frac{x}{y}+\frac{y}{x} = 34 ,$</p>
</li>
<li>
<p>$\log_am = \log_an, \Leftrightarrow m=n$</p>
</li>
</ul>
</div>
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</div>
</main>
<hr>
<footer style = "font-size: 18px">Problem 2 $\rightarrow$ Problem 2 $\rightarrow$ <span style = "color:blue"> Problem 3 </span> $\rightarrow$ Problem 4 $\rightarrow$ Problem 5 </footer>

<h3 id="success-stylefont-size25px-logarithm-l-3-success-7"><Success style="font-size:25px"> Logarithm L-3 </Success></h3>
<h3 id="primary-stylefont-size24px-problem-4-primary"><primary style="font-size:24px"> Problem 4 </primary></h3>
<hr>
<main>
<div style='float: left; width:99%;font-size:25px; text-align:left;'>
<ul>
<li>
<ol start="4">
<li>If $a^2-12 a b+4 b^2=0$.</li>
</ol>
</li>
<li>
<p>Prove that  $ \quad \log (a+2 b)=$</p>
</li>
<li>
<p>$\frac{1}{2}(\log a+\log b)+2 \log 2$</p>
</li>
<li>
<p>Solution.  Given$  \quad a^2-12 a b+4 b^2=0$</p>
</li>
<li>
<p>To show. $\log (a+2 b)=\frac{1}{2}(\log a+\log b)+2 \log 2$</p>
</li>
<li>
<p>$\frac{1}{2} $ Power law,  $(\log a+\log b)$ Multiplication law,  2 $\log 2$ power law</p>
</li>
<li>
<p>$\log ()=\log (\quad)$</p>
</li>
<li>
<p>$\log (a+2 b)  =\log (a b)^{1 / 2}+\log 2^2 $</p>
</li>
<li>
<p>$\log (a+2 b)  =\log (4 \sqrt{a b}) $</p>
</li>
<li>
<p>$(a+2 b)  =4 \sqrt{a b} \quad \rightarrow (a+2 b)^2  =16 a b $</p>
</li>
<li>
<p>$a^2+4 b^2+4 a b  =16 a b \Rightarrow a^2+4 b^2-12 a b  =0$</p>
</li>
</ul>
</div>
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</main>
<hr>
<footer style = "font-size: 18px">Problem 2 $\rightarrow$ Problem 3 $\rightarrow$ <span style = "color:blue"> Problem 4 </span> $\rightarrow$ Problem 5 $\rightarrow$ Problem 6 </footer>

### <Success style="font-size:25px"> Logarithm L-3 </Success>
### <primary style="font-size:24px"> Problem 5 </primary>
<hr>
<main>

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

- Soln.  <span style="color:green">$ \frac{2}{3}<\log _{10} 5 $</span>

- $\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 $

- $m<n \Leftrightarrow \log _a m<\log _a n$
 
- <span style="color:green">$\log _{10} 5<\frac{3}{4} $ </span>

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

</div>
</main>
<hr>
<footer style = "font-size: 18px">Problem 3 $\rightarrow$ Problem 4 $\rightarrow$ <span style = "color:blue"> Problem 5 </span> $\rightarrow$ Problem 6 $\rightarrow$ Problem 7 </footer>

### <Success style="font-size:25px"> Logarithm L-3 </Success>
### <primary style="font-size:24px"> Problem 6 </primary>
<hr>
<main>

- 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) $

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

</div>
</main>
<hr>
<footer style = "font-size: 18px">Problem 4 $\rightarrow$ Problem 5 $\rightarrow$ <span style = "color:blue"> Problem 6 </span> $\rightarrow$ Problem 7 $\rightarrow$ Problem 7 </footer>

<h3 id="success-stylefont-size25px-logarithm-l-3-success-8"><Success style="font-size:25px"> Logarithm L-3 </Success></h3>
<h3 id="primary-stylefont-size24px-problem-7-primary"><primary style="font-size:24px"> Problem 7 </primary></h3>
<hr>
<main>
<div style='float: left; width:99%;font-size:25px; text-align:left;'>
<ul>
<li>
<p>Solve for $x, \log _4(x-1)=\log _2(x-3)$</p>
</li>
<li>
<p>Soln.  $2^2=4,4^{1 / 2}=2$</p>
</li>
<li>
<p>$ \log _a b=\frac{\log _c b}{\log _c a} $</p>
</li>
<li>
<p>RHS $=\log _2(x-3)=\log _{4^{1 / 2}}(x-3) $</p>
</li>
<li>
<p>$=\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) $</p>
</li>
<li>
<p>$\log _4(x-1)  =\log _4(x-3)^2$</p>
</li>
<li>
<p>$\Leftrightarrow \quad(x-1)  =(x-3)^2 $</p>
</li>
<li>
<p>$x-1  =x^2-6 x+9 $</p>
</li>
<li>
<p>$x^2-7 x  +10=0$</p>
</li>
<li>
<p>$ \Leftrightarrow(x-5)(x-2)=0$</p>
</li>
<li>
<p>$x=5,2$   are the possible solutions</p>
</li>
</ul>
</div>
</main>
<hr>
<footer style = "font-size: 18px">Problem 5 $\rightarrow$ Problem 6 $\rightarrow$ <span style = "color:blue"> Problem 7 </span> $\rightarrow$ Problem 7 $\rightarrow$ Thankyou </footer>

<h3 id="success-stylefont-size25px-logarithm-l-3-success-9"><Success style="font-size:25px"> Logarithm L-3 </Success></h3>
<h3 id="primary-stylefont-size24px-problem-7-primary-1"><primary style="font-size:24px"> Problem 7 </primary></h3>
<hr>
<main>
<div style='float: left; width:99%;font-size:25px; text-align:left;'>
<ul>
<li>
<p>Case 1.$ x=2 \quad \log (x-3)=\log \sqrt{4} (-1) $</p>
</li>
<li>
<p>$x=5 \quad \log _2(x-3)=\log _2 2=1$</p>
</li>
<li>
<p>$x=5$</p>
</li>
</ul>
</div>
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</main>
<hr>
<footer style = "font-size: 18px">Problem 6 $\rightarrow$ Problem 7 $\rightarrow$ <span style = "color:blue"> Problem 7 </span> $\rightarrow$ Thank you $\rightarrow$  </footer>