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Logarithm Formula Sheet

This page will help you to revise formulas and concepts of Logarithm instantly for various exams.
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Logarithm is a mathematical function that helps to find the exponent (or power) to which a base number must be raised to obtain a given value.

Neetesh Kumar | May 10, 2024                                       \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space Share this Page on: Reddit icon Discord icon Email icon WhatsApp icon Telegram icon

1. Definition:

The logarithm of the number N to the base ‘a’ is the exponent indicating the power to which the base ‘a’ must be raised to obtain the number N. This number is designated as loga_aN.

2. Properties and Formulas:

(a)\bold{(a)} loga_aN = x, read as log of N to the base a     \iff ax=Na^x = N.
If a = 10, then we write log N or log10_{10}N and if a = e, we write ln N or lne_eN (Natural log)

(b)\bold{(b)} Necessary conditions : N > 0 ; a > 0 ; a \ne 1

(c)\bold{(c)} loga_a1 = 0

(d)\bold{(d)} loga_aa = 1

(e)\bold{(e)} log1a_\frac{1}{a}a = -1

(f)\bold{(f)} loga_ax.y = loga_ax + loga_ay; x, y > 0

(g)\bold{(g)} loga(xy)_a(\frac{x}{y}) = loga_ax - loga_ay; x, y > 0

(h)\bold{(h)} logaxp_ax^p = ploga_ax ; x > 0

(i)\bold{(i)} logaq_{a^q}x = 1q\frac{1}{q}loga_ax ; x > 0

(j)\bold{(j)} loga_ax = 1logax\frac{1}{log_ax} ; x > 0, x \ne 1

(k)\bold{(k)} loga_ax = logaxlogbx\frac{log_ax}{log_bx} ; x > 0, a,b > 0, a, b \ne 1

(l)\bold{(l)} loga_ab.logb_bc.logc_cd = loga_ad; a, b, c, d > 0, \ne 1

(m)\bold{(m)} alogaxa^{log_ax} = x, a > 0, a \ne 1

(n)\bold{(n)} alogbca^{log_bc} = clogba;c^{log_ba}; a, b, c > 0, b \ne 1

(o)\bold{(o)} loga_ax < loga_ay     \iff x = {x<yif a>1x>yif 0<a<1\begin{cases} x < y &\text{if } a > 1 \\ x > y &\text{if } 0 < a < 1 \end{cases}

(p)\bold{(p)} loga_ax = loga_ay \Rightarrow x = y; x, y > 0; a > 0, a \ne 1

(q)\bold{(q)} elnax=axe^{lna^x} = a^x

(r)\bold{(r)} log10_{10}2 = 0.3010, log10_{10}3 = 0.4771, ln2 = 0.693, ln10 = 2.303

(s)\bold{(s)} If a > 1 then loga_ax < p \Rightarrow 0 < x < apa^p

(t)\bold{(t)} If a > 1 then loga_ax > p \Rightarrow x > apa^p

(u)\bold{(u)} If 0 < a < 1 then loga_ax < p \Rightarrow x > apa^p

(v)\bold{(v)} If 0 < a < 1 then loga_ax > p \Rightarrow 0 < x < apa^p

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