Number theory - Logarithms/Modulus function

Logarithms

$$a^x = N \iff x = \log_a N , \quad a > 0, \, a \neq 1, \, N > 0$$

LHS is exponential form while the RHS is same thing written in logarithmic form.

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NOTE:

(i) $\log_a 1 = 0$ 

(ii) $\log_a a = 1$

(iii) $\log_a (\text{-ve})$ is not defined

(iv) $\log_a 0$ is not defined

Properties:

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1.

$$\log_a(mn) = \log_a m + \log_a n$$

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In general,

$${\log }_a(x_1\cdot x_2\cdot x_3\cdots x_n)={\log }_a{x}_1+{\log }_a{x}_2+{\log }_a{x}_3+\cdots +{\log }_a{x}_n$$

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2.

$$\log_a\left(\frac{m}{n}\right) = \log_a m - \log_a n$$

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3.

$$\log_a(m^n) = n \log_a m$$

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4.

$$\log_{a^p}(m^n) = \frac{n}{p} \log_a m$$

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5.

$$\log_b a = \frac{1}{\log_a b}$$

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6.

$$\log_b a = \frac{\log a}{\log b}$$

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7.

$$\log_b a \cdot \log_c b \cdot \log_d c = \log_d a$$

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8.

$$a^{\log_a N} = N$$

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9.

 $a^{\log_b c}=c^{\log_b a}$.

Quick proof:
Let $a=b^x$ and $c=b^y$ where $x=\log_b a$ and $y=\log_b c$.
Then $(b^x)^y=b^{xy}=(b^y)^x$, so $a^{\log_b c}=c^{\log_b a}$.

Ex1.

Find ‘x’, if $9^{\log_{3} x} = 16$

Sol –
$(3^2)^{\log_{3} x} = 16$

$3^{2 \log_{3} x} = 16$

$3^{\log_{3} x^2} = 16 \Rightarrow x^2 = 16$

$\Rightarrow x=\pm 4$

But x can't be negative so x = 4 = Answer.

Ex2.

2. Simplify $\left(\frac{1}{2}\right)^{\log_{2} 6}$

Answer: 1/6

Ex3.

3. $\log_{2} 3 \cdot \log_{3} 4 \cdot \log_{4} 5 \cdots \log_{511} 512$

Answer: 9

Ex4.

4. Find the value of ‘x’, if $5^{2 \log_{5} x} - x - 6 = 0$

Sol –
$x^2 - x - 6 = 0$

$(x - 3)(x + 2) = 0$

$x=3,-2$

But x can't be negative, so answer: x = 3.

Logarithmic inequalities:
If base 0 < a < 1 then the sign flips, else it remains the same.
With base > 1:

$\log_{3} x < 5$

$x < 3^5$

$x<243$

With base < 1:

$\log_{0.1} x > 10$

$x < (0.1)^{10}$

$x<\frac{1}{{10}^{10}}$

Also:

$\log_{0.1} 10$

$\log_{10} \left(\frac{1}{10}\right)$

$\log_{10} 10^{-1}$

$= -1 \cdot (\log_{10} 10)$

$= -1$

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$\log_{0.1} 100$

$\log_{10} 100^{-1}$

$-2 \cdot (\log_{10} 10)$

$= -2$

$10<100$

But their logs base 0.1 are inverted in comparison.

Ex1.

Find interval of ‘x’ satisfying
${\log }_{\frac{1}{3}}\left(\frac{2-3x}{x}\right)\ge -1$

Solution:

$$\log_{1/3}\!\left(\frac{2-3x}{x}\right)\ge -1.$$

Since the base $1/3\in(0,1)$, $\log_{1/3}$ , so

$$\log_{1/3} y \ge -1 \iff y \le (1/3)^{-1}=3,$$

also, log of a number is defined only when it's > 0 So

$$\frac{2-3x}{x}>0 \quad\text{and}\quad \frac{2-3x}{x}\le 3.$$

Domain: $\frac{2-3x}{x}>0 \Rightarrow 0<x<\frac{2}{3}$.

We got this using wavy curve method.

On this interval $x>0$, so multiply by $x$ without flipping the sign:

$$2-3x\le 3x \;\Rightarrow\; 2\le 6x \;\Rightarrow\; x\ge \frac{1}{3}.$$

Intersection: $\boxed{\;x\in\left[\frac{1}{3},\,\frac{2}{3}\right)\;}$.

(Check: at $x=\tfrac13$, the log equals $-1$; at $x=\tfrac23$ the argument is $0$, not allowed.)

Modulus function properties:

For any $a, b \in \mathbb{R}$,

(i) $|a| \geq 0$

(ii) $|a| = |-a|$

(iii) $|a \cdot b| = |a| \cdot |b|$

(iv) $\left|\frac{a}{b}\right| = \frac{|a|}{|b|}, \quad (b \neq 0)$

(v) $\mathrm{\mid }a+b\mathrm{\mid }\le \mathrm{\mid }a\mathrm{\mid }+\mathrm{\mid }b\mathrm{\mid }$

(vi) $|a| - |b| \leq |a - b|$

(vii) $\big||a| - |b|\big| = |a - b|$ iff $ab\ge 0$

If $a > 0$,

$\mathrm{\mid }x\mathrm{\mid }<a\Rightarrow -a<x<a$

$\mathrm{\mid }x\mathrm{\mid }>a\Rightarrow x<-a\text{ or }x>a$

Ex 1.

Solve |2x + 1| < 7

So: -7 < 2x + 1 < 7

-7 - 1 < 2x + 1 - 1 < 7 - 1

-8 < 2x < 6

-4 < x < 3

x ∈ (-4, 3)

Ex2.

$|x - 3| + |x + 4| \geq 10$

Critical points: $3, -4$

$x<-4;-4\le x\le 3;x>3$

We need to test it in above 3 intervals.

Case 1: When x < -4, both terms inside modulus are negative so we can write:
-(x-3) - (x+4) >= 10
=> -2x-1 >= 10
=> -2x >= 11
=> x <= -11/2 which is indeed < -4 so valid

Case 2: -4 <= x <= 3 => 
First term is negative but second is positive.
-(x-3) + (x+4) >= 10
=> 7 >= 10 Invalid

Case 3: x > 3
Both terms are positive.
x - 3 + x + 4 >= 10
=> 2x >= 9 => x >= 9/2 which complies with x > 3. 
Answer: x <= -11/2 Union x > 3.