JEE Mathematics Logarithms Notes Set 04

Definition and Domain

\( \log_aN=x\iff a^x=N \), valid only when \( a\gt0,\ a\ne1,\ N\gt0 \). The base cannot be 1 because every power of 1 is 1. Common logs use base 10 and natural logs use base \( e\approx2.718 \). Remember \( \log_{10}2\approx0.3010 \), \( \log_{10}3\approx0.4771 \), \( \ln2\approx0.693 \) and \( \ln10\approx2.303 \).
The graph of \( y=\log_ax \) rises for \( a\gt1 \) and falls for \( 0\lt a\lt1 \), always passing through \( (1,0) \). It is the mirror image of \( y=a^x \) in the line \( y=x \).
Sign rule: if the number and the base lie on the same side of 1 the logarithm is positive; if on opposite sides it is negative.

Laws of Logarithms

\( \log_a(MN)=\log_aM+\log_aN \), \( \log_a\frac MN=\log_aM-\log_aN \), \( \log_aM^p=p\log_aM \), \( \log_{a^q}M=\frac1q\log_aM \).
Change of base: \( \log_bM=\dfrac{\log_aM}{\log_ab} \), which gives \( \log_ba=\dfrac1{\log_ab} \) and \( \log_ab\cdot\log_bc=\log_ac \).
Fundamental identity: \( a^{\log_aN}=N \). Also \( a^{\log_cb}=b^{\log_ca} \), \( \log_a1=0 \), \( \log_aa=1 \), \( \log_{1/a}a=-1 \).
A product of consecutive logs often telescopes: \( \log_23\cdot\log_34\cdots\log_{n-1}n=\log_2n \).

Logarithmic Equations

Always write down the domain first (arguments positive, base positive and not 1). Then reduce to one of the standard forms: \( \log_ax=\log_ay\iff x=y \), or convert \( \log_af(x)=c \) to \( f(x)=a^c \). Equations in several logarithms are often quadratic in \( t=\log_ax \); for example \( t+\frac1t=\frac52 \) arises from \( \log_3x+\log_x3=\frac52 \).
Finally reject every root that violates the domain. Extra roots appear whenever a step such as combining \( \log(x)+\log(x-2) \) widens the domain.

Logarithmic Inequalities

For base \( a\gt1 \) the logarithm is increasing: \( \log_ax\gt\log_ay\iff x\gt y\gt0 \). For base \( 0\lt a\lt1 \) it is decreasing: \( \log_ax\gt\log_ay\iff 0\lt x\lt y \). The inequality reverses only in the second case, and the arguments must stay positive in both cases.
With a variable base \( g(x) \), split into the two cases \( g(x)\gt1 \) and \( 0\lt g(x)\lt1 \): \( \log_{g}f\ge a \) means \( f\ge g^a \) when \( g\gt1 \) and \( 0\lt f\le g^a \) when \( 0\lt g\lt1 \). Solve each case and take the union.
Exponential inequalities such as \( a^{f(x)}\gt a^{g(x)} \) follow the same rule: keep the sign for \( a\gt1 \) and reverse it for \( 0\lt a\lt1 \).

Characteristic and Mantissa

For base 10 write \( \log_{10}N=n+f \) where the integer \( n=[\log_{10}N] \) is the characteristic and \( f=\{\log_{10}N\}\in[0,1) \) is the mantissa. If the characteristic is \( n\ge0 \), the number \( N \) has \( n+1 \) digits before the decimal point.
If the characteristic is \( -k \) (\( k\ge1 \)), then \( N \) is less than 1 and has \( k-1 \) zeros immediately after the decimal point before the first significant digit. For example \( \log_{10}0.0004=-3.398=\overline4.602 \), so there are 3 zeros.

Example 1. Solve \( \log_2x+\log_2(x-2)=3 \).
Answer:
Domain: \( x\gt2 \).
Combine: \( x(x-2)=8 \), so \( x^2-2x-8=0 \) and \( x=4 \) or \( -2 \).
\( x=4 \) (the root \( -2 \) is outside the domain).

Example 2. Solve \( \log_{1/3}(2x-1)\ge-1 \).
Answer:
Base lies between 0 and 1, so the inequality reverses for the arguments: \( 0\lt2x-1\le(1/3)^{-1}=3 \).
This gives \( \frac12\lt x\le2 \).
\( x\in\left(\tfrac12,2\right] \)

Example 3. How many digits does \( 3^{40} \) have? (Use \( \log_{10}3=0.4771 \).)
Answer:
\( \log_{10}3^{40}=40\times0.4771=19.084 \), so the characteristic is 19.
It has 20 digits.

Example 4. Solve \( \log_2(x^2-3x)\lt2 \).
Answer:
Domain: \( x^2-3x\gt0 \), i.e. \( x\lt0 \) or \( x\gt3 \). Inequality: \( x^2-3x\lt4 \), i.e. \( -1\lt x\lt4 \).
\( x\in(-1,0)\cup(3,4) \)