JEE Mathematics Logarithms Notes Set 02

What a Logarithm Tells You

1. \(\log_aN\) is the power to which the base \(a\) must be raised to give \(N\). Writing \(\log_aN=x\) is the same as writing \(a^x=N\), and it makes sense only for \(N>0\), \(a>0\), \(a\ne1\).
2. Two special bases are used: \(\log\) without a base means base 10 (common logarithm) and \(\ln\) means base \(e\) (natural logarithm). In this chapter \(\log x\) is taken as \(\log_{10}x\).
3. Fundamental identity: \(a^{\log_aN}=N\). Standard values: \(\log_a1=0\), \(\log_aa=1\), \(\log_{1/a}a=-1\).
4. Sign rule: when the base and the number are on the same side of 1 the logarithm is positive; when they are on opposite sides it is negative. This is why \(\log_x(a^2+1)<0\) forces \(0<x<1\).

Rules You Will Use in Every Problem

A very useful habit: when several logarithms to different bases appear, convert them all to one convenient base, such as 2 or 10, or to a single variable such as \(t=\log_2 3\). A product of chained logarithms such as \(\log_34\cdot\log_45\cdots\) collapses to one logarithm, and a sum like \(\dfrac1{\log_2n}+\dfrac1{\log_3n}+\cdots\) collapses to \(\log_n\) of a product.

Equations With Logarithms

1. Always write the domain first: each argument positive, each variable base positive and not equal to 1.
2. If both sides are single logarithms to the same base, equate the arguments. If one side is a number, write the number as a logarithm of the same base.
3. If a logarithm appears squared or under a root, substitute \(t=\log_ax\) (or \(t=\sqrt{\log_ax}\)) and solve the quadratic in \(t\).
4. For equations such as \(x^{\log x}=10x^2\) or \(x^{f(x)}=c\), take the logarithm of both sides to turn the exponent into a multiplier.
5. For equations with \(3^x\) or \(4^x\) inside a logarithm, put \(y=3^x\) and remember that \(y>0\).
6. When a root of an equation makes a denominator zero (for example a logarithm equal to 0 below a fraction), reject it. Roots found after squaring must be verified in the original equation.

Inequalities

1. For base greater than 1, \(\log_ax<\log_ay\) means \(0<x<y\); the direction is preserved.
2. For base between 0 and 1, the direction reverses: \(\log_ax<\log_ay\) means \(x>y>0\).
3. For \(a>1\): \(\log_ax<p\Rightarrow0<x<a^p\) and \(\log_ax>p\Rightarrow x>a^p\). For \(0<a<1\): \(\log_ax<p\Rightarrow x>a^p\) and \(\log_ax>p\Rightarrow0<x<a^p\).
4. Whenever the base is a variable, split the problem into two cases: base greater than 1 and base between 0 and 1.

Characteristic, Mantissa and Number of Digits

1. Write \(\log_{10}N=\) integer \(+\) fraction. The integer part is the characteristic and the fractional part \(0\le\beta<1\) is the mantissa; the mantissa is never negative.
2. For \(N\ge1\), the number of digits of N is the characteristic plus 1.
3. For \(0<N<1\) with characteristic \(-k\), there are \(k-1\) zeros after the decimal point before the first non-zero digit.
4. A negative logarithm such as \(-16.48\) is written as \(-17+0.52\), so the characteristic is \(-17\) and the mantissa is \(0.52\). Shifting the decimal point of a number changes only the characteristic, not the mantissa.
5. In another base b, the numbers with characteristic k satisfy \(b^k\le N<b^{k+1}\), so there are \(b^{k+1}-b^k\) such integers.

Worked Examples

1. If \(\log_{12}27=a\), find \(\log_616\). Put \(p=\log_23\): \(a=\dfrac{3p}{2+p}\), so \(p=\dfrac{2a}{3-a}\). Then \(\log_616=\dfrac4{1+p}=\dfrac{4(3-a)}{3+a}\).
2. Solve \(\log_2(3x-2)=\log_{1/2}x\). Since \(\log_{1/2}x=-\log_2x\), \(3x-2=\dfrac1x\), giving \(x=1\) or \(-\dfrac13\). The domain \(x>\dfrac23\) leaves \(x=1\).
3. How many digits does \(2^{75}\) have? \(75\log2=22.575\), characteristic 22, so 23 digits.
4. Solve \(x^{\log x+1}=10^6\). Take \(\log\): \(t(t+1)=6\) with \(t=\log x\), so \(t=2\) or \(-3\) and \(x=100\) or \(10^{-3}\).
5. Simplify \(\log_z\)(expression) when \(x^6-y^6=z^6\): factorise the arguments with the sum and difference of cubes; the product of the arguments is \(x^6-y^6\), so the logarithm equals 6.