Theory of Logarithm
Definition: A number $x$ is called the logartihm of a number $y$ to the base $b$ if $b^x = y$ where $b>0, b\neq , y > 0$
Mathematically, it is represented by the equation $\log_b y = x or b^x = y$
Notes:
- The conditions $b>0, b\neq 1$ and $y > 0$ are necessary in the definition of logarithm.
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When $b=1$ suppose logarithm is defined, and we have to find the value of $\log_1y = x.$ Let $\log_1y = x\Rightarrow 1^x = y \Rightarrow 1 = y$
If $\log_12$ is defined then $1=2.$ So we see that $b=1$ leads to meaningless result.
- Similalrly if $y<0$, then $b^x = y$, which is meaningless as L.H.S. is positive and R.H.S. is negative.
- Let the condition to be true when $b=0.$ Thus, $0^x = N$ i.e. if $\log_02$ is defined will mean that $0=2$, which signifies that our assumption is false.
- No number can have two different logarithms to a given base. Assume that a number $N$ has two different logarithms $x$ and $y$ with base $b.$ Then, $\log_b N = x, \log_bN = y\Rightarrow N = b^x, N = b^y \Rightarrow b^x = b^y \Rightarrow x = y$.
- When the number or base is negative the value of logarithm comes out to be a complex number with non-zero imaginary part. Let $\log_e(-5) = x\Rightarrow \log_e5.e^{i\pi} = x\Rightarrow x = \log_e5 + i\pi$
Important Results
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$\log_b 1 = 0$
Proof: Let $\log_b 1= x\Rightarrow b^x = 1\Rightarrow x = 0$ -
$\log_bb = 1$
Proof: Let $\log_bb = x\Rightarrow b^x = b \Rightarrow x = 1$ -
$b^{\log_bN} = N$
Proof: Let $\log_bN = x \Rightarrow b^x = N$ $\Rightarrow b^{\log_bN} = N$
Important Formulas
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$\log_b(x.y) = \log_b x + \log_by(x > 0, y >0)$
Proof: Let $\log_bx = m\Rightarrow b^m = x$ and $\log_by = n \Rightarrow b^n = y$
$x.y = b^m.b^n = b^{m+n} = b^0$(say) $\Rightarrow m + n = 0$
$\log_bx.y = \log_b x + \log_b y$
Corollary: $\log_b(x.y.z) = \log_bx + \log_by + \log_bz$
If $x<0, y<0, \log_b(x.y)=\log_b|x| + \log_b|y|$ -
$\log_b\left(\frac{x}{y}\right)=\log_b x - \log_by(x>0, y>0)$
Proof: Let $\log_bx = m \Rightarrow b^m = x$ and $\log_by = n \Rightarrow b^n = y$
$\log_b\left(\frac{x}{y}\right) = o \Rightarrow b^o = \frac{x}{y}$
$\frac{x}{y} = \frac{b^m}{b^n} = b^{m - n} = b^o \Rightarrow m - n = o$
$\log_b\left(\frac{x}{y}\right) = \log_b|x| - \log_b|y|(x<0, y<0)$