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Toolkit 48

Logarithms: Definition and Properties

Definition

log⁡bx=y  ⟺  by=x\log_b x = y \iff b^y = x

Domain

x>0,b>0,b≠1x>0,\qquad b>0,\qquad b\ne 1

Range

(−∞,+∞)(-\infty,+\infty)

If b>1b>1, then y=log⁡bxy=\log_b x is increasing.

If 0<b<10<b<1, then y=log⁡bxy=\log_b x is decreasing.

Graph of y = log_b(x) for b>1 and 0<b<1, showing the increasing and decreasing cases.

Properties

1.log⁡b1=01.\quad \log_b 1 = 0
2.log⁡bb=12.\quad \log_b b = 1
3.log⁡b(bm)=m3.\quad \log_b(b^m) = m
4.blog⁡ba=a4.\quad b^{\log_b a} = a
5.log⁡ba+log⁡bc=log⁡b(ac)5.\quad \log_b a + \log_b c = \log_b(ac)
6.log⁡ba−log⁡bc=log⁡b ⁣(ac)6.\quad \log_b a - \log_b c = \log_b\!\left(\frac{a}{c}\right)
7.log⁡b(an)=nlog⁡ba7.\quad \log_b(a^n) = n\log_b a
8.log⁡bma=1mlog⁡ba8.\quad \log_{b^m} a = \frac{1}{m}\log_b a
9.log⁡b ⁣(1a)=−log⁡ba9.\quad \log_b\!\left(\frac{1}{a}\right) = -\log_b a
10.log⁡ba⋅log⁡ac=log⁡bc10.\quad \log_b a \cdot \log_a c = \log_b c
11.log⁡ba=log⁡calog⁡cb11.\quad \log_b a = \frac{\log_c a}{\log_c b}
12.log⁡ba=ln⁡aln⁡b,where ln⁡x=log⁡ex, e=2.71828…12.\quad \log_b a = \frac{\ln a}{\ln b},\quad \text{where } \ln x = \log_e x,\ e=2.71828\ldots
13.log⁡ba=1log⁡ab13.\quad \log_b a = \frac{1}{\log_a b}
14.b(log⁡ax)(log⁡ba)=x14.\quad b^{(\log_a x)(\log_b a)} = x
15.log⁡bx=log⁡by  ⟺  x=y15.\quad \log_b x = \log_b y \iff x = y

Proof

1.

log⁡b1=0\log_b 1 = 0

Proof. By the definition of a logarithm, log⁡b1=0  ⟺  b0=1.\log_b 1 = 0 \iff b^0 = 1.

2.

log⁡bb=1\log_b b = 1

Proof. By the definition of a logarithm, log⁡bb=1  ⟺  b1=b.\log_b b = 1 \iff b^1 = b.

3.

log⁡b(bm)=m\log_b(b^m) = m

Proof. By the definition of a logarithm, log⁡b(bm)=m  ⟺  bm=bm.\log_b(b^m) = m \iff b^m = b^m.

4.

blog⁡ba=ab^{\log_b a} = a

Proof.

Let

log⁡ba=m.\log_b a = m.

Then

bm=a,b^m = a,

so

blog⁡ba=a.b^{\log_b a} = a.

5.

log⁡ba+log⁡bc=log⁡b(ac)\log_b a + \log_b c = \log_b(ac)

Proof.

Let

log⁡ba=m,log⁡bc=n.\log_b a = m, \qquad \log_b c = n.

Then

a=bm,c=bn.a = b^m, \qquad c = b^n.

Therefore,

ac=bm+n,ac = b^{m+n},

which implies

log⁡b(ac)=m+n=log⁡ba+log⁡bc.\log_b(ac) = m+n = \log_b a + \log_b c.

6.

log⁡ba−log⁡bc=log⁡b ⁣(ac)\log_b a - \log_b c = \log_b\!\left(\frac{a}{c}\right)

Proof.

Let

log⁡ba=m,log⁡bc=n.\log_b a = m, \qquad \log_b c = n.

Then

a=bm,c=bn.a = b^m, \qquad c = b^n.

Hence

ac=bm−n,\frac{a}{c} = b^{m-n},

so

log⁡b ⁣(ac)=m−n=log⁡ba−log⁡bc.\log_b\!\left(\frac{a}{c}\right) = m-n = \log_b a - \log_b c.

7.

log⁡b(an)=nlog⁡ba\log_b(a^n) = n\log_b a

Proof.

Let

log⁡ba=m.\log_b a = m.

Then

a=bm,a = b^m,

so

an=bnm.a^n = b^{nm}.

Therefore,

log⁡b(an)=nm=nlog⁡ba.\log_b(a^n) = nm = n\log_b a.

8.

log⁡bma=1mlog⁡ba\log_{b^m} a = \frac{1}{m}\log_b a

Proof.

Let

log⁡ba=n.\log_b a = n.

Then

a=bn=(bm) n/m.a = b^n = (b^m)^{\,n/m}.

Hence

log⁡bma=nm=1mlog⁡ba.□\log_{b^m} a = \frac{n}{m} = \frac{1}{m}\log_b a. \quad\square

9.

log⁡b ⁣(1a)=−log⁡ba\log_b\!\left(\frac{1}{a}\right) = -\log_b a

Proof.

Let

log⁡ba=m.\log_b a = m.

Then

a=bm,a = b^m,

so

1a=b−m.\frac{1}{a} = b^{-m}.

Therefore,

log⁡b ⁣(1a)=−m=−log⁡ba.\log_b\!\left(\frac{1}{a}\right) = -m = -\log_b a.

10.

log⁡ba⋅log⁡ac=log⁡bc\log_b a \cdot \log_a c = \log_b c

Proof.

Let

log⁡ba=m.\log_b a = m.

Then

a=bm.a = b^m.

Also let

log⁡ac=n.\log_a c = n.

Then

c=an=(bm)n=bmn.c = a^n = (b^m)^n = b^{mn}.

Therefore,

log⁡bc=mn=log⁡ba⋅log⁡ac.\log_b c = mn = \log_b a \cdot \log_a c.

11.

log⁡ba=log⁡calog⁡cb\log_b a = \frac{\log_c a}{\log_c b}

Proof.

Using Property 10,

log⁡ba⋅log⁡cb=log⁡ca.\log_b a \cdot \log_c b = \log_c a.

Dividing both sides by log⁡cb\log_c b,

log⁡ba=log⁡calog⁡cb.\log_b a = \frac{\log_c a}{\log_c b}.

12.

log⁡ba=ln⁡aln⁡b\log_b a = \frac{\ln a}{\ln b}

where ln⁡x=log⁡ex.\ln x = \log_e x.

Proof.

Applying Property 10,

log⁡ba⋅ln⁡b=log⁡ba⋅log⁡eb=ln⁡a.\log_b a \cdot \ln b = \log_b a \cdot \log_e b = \ln a.

Hence,

log⁡ba=ln⁡aln⁡b.\log_b a = \frac{\ln a}{\ln b}.

13.

log⁡ba=1log⁡ab\log_b a = \frac{1}{\log_a b}

Proof.

Using Properties 10 and 2,

log⁡ba⋅log⁡ab=log⁡bb=1.\log_b a \cdot \log_a b = \log_b b = 1.

Therefore,

log⁡ba=1log⁡ab.\log_b a = \frac{1}{\log_a b}.

14.

b(log⁡ax)(log⁡ba)=xb^{(\log_a x)(\log_b a)} = x

Proof.

Using Properties 4 and 10,

b(log⁡ax)(log⁡ba)=(blog⁡ba)log⁡ax=alog⁡ax=x.b^{(\log_a x)(\log_b a)} = \left(b^{\log_b a}\right)^{\log_a x} = a^{\log_a x} = x.

15.

log⁡bx=log⁡by  ⟺  x=y\log_b x = \log_b y \iff x = y

Proof.

If

log⁡bx=log⁡by,\log_b x = \log_b y,

then

blog⁡bx=blog⁡by,b^{\log_b x} = b^{\log_b y},

which implies

x=y.x = y.

Conversely, if

x=y,x = y,

then

log⁡bx=log⁡by.\log_b x = \log_b y.

Thus,

log⁡bx=log⁡by  ⟺  x=y.□\log_b x = \log_b y \iff x = y. \quad\square