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Logarithmic Functions

Understand logarithmic graphs, domain restrictions, and change of base. Three worked examples cover comparisons, simplification, and a variable base.

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Tutorial Content

Logarithmic Functions

A logarithm answers a question about an exponent. For example, log⁡28=3\log_2 8=3 because 23=82^3=8. This connection lets you move between logarithmic and exponential equations:

y=log⁡ax⟺ay=x.y=\log_a x \quad\Longleftrightarrow\quad a^y=x.

For real logarithms, the base must satisfy a>0a>0 and a≠1a\ne1, and the argument must be positive. In this tutorial, lg⁡x\lg x means log⁡10x\log_{10}x, while ln⁡x\ln x means log⁡ex\log_e x.

1. Reading the graph

The parent function y=log⁡axy=\log_a x has domain (0,+∞)(0,+\infty) and range R\mathbb{R}. It passes through (1,0)(1,0) because a0=1a^0=1, and through (a,1)(a,1) because a1=aa^1=a. The line x=0x=0 is a vertical asymptote; the graph never meets the yy-axis.

Its direction depends on the base:

  • If a>1a>1, the function is increasing.
  • If 0<a<10<a<1, the function is decreasing.

The following exact values show the difference without relying on a sketch.

xxlog⁡2x\log_2 xlog⁡1/2x\log_{1/2}x
12\frac12−1-111
110000
2211−1-1
4422−2-2

For a>1a>1, log⁡ax\log_a x tends to −∞-\infty as x→0+x\to0^+ and to +∞+\infty as x→+∞x\to+\infty. For 0<a<10<a<1, these limits are reversed.

The functions y=log⁡axy=\log_a x and y=axy=a^x are inverses. Their graphs are reflections in the line y=xy=x: a point (u,v)(u,v) on one corresponds to (v,u)(v,u) on the other.

These properties describe the parent function. For an expression such as log⁡a(x−2)\log_a(x-2), check the whole argument: the domain is x>2x>2, and the vertical asymptote is x=2x=2.

2. Using logarithm laws

For M>0M>0, N>0N>0, a>0a>0, a≠1a\ne1, and a real number rr,

log⁡a(MN)=log⁡aM+log⁡aN,\log_a(MN)=\log_a M+\log_a N, log⁡a ⁣(MN)=log⁡aM−log⁡aN,\log_a\!\left(\frac{M}{N}\right)=\log_a M-\log_a N, log⁡a(Mr)=rlog⁡aM.\log_a(M^r)=r\log_a M.

The positivity conditions matter. For example, log⁡a(x2)\log_a(x^2) is defined when x≠0x\ne0, but 2log⁡ax2\log_a x is defined only when x>0x>0. An identity valid for every nonzero real xx is

log⁡a(x2)=2log⁡a∣x∣.\log_a(x^2)=2\log_a|x|.

There is no general rule that splits a logarithm of a sum. Taking M=N=1M=N=1 gives log⁡a(M+N)=log⁡a2\log_a(M+N)=\log_a2, whereas log⁡aM+log⁡aN=0\log_a M+\log_a N=0.

3. Changing the base

To express a logarithm in a convenient base bb, use

log⁡ax=log⁡bxlog⁡ba,a>0, a≠1, b>0, b≠1, x>0.\log_a x=\frac{\log_b x}{\log_b a}, \qquad a>0,\ a\ne1,\ b>0,\ b\ne1,\ x>0.

Two useful consequences are

log⁡ab  log⁡bc=log⁡ac\log_a b\;\log_b c=\log_a c

for valid bases a,ba,b and c>0c>0, and

log⁡an(bm)=mnlog⁡ab\log_{a^n}(b^m)=\frac{m}{n}\log_a b

for a>0a>0, a≠1a\ne1, b>0b>0, real mm, and real n≠0n\ne0. The last condition is necessary: n=0n=0 would make the base equal to 11.

4. Worked examples

Example 1: Compare logarithms with the same base

Compare log⁡23\log_2 3 with log⁡25\log_2 5, then compare log⁡0.53\log_{0.5}3 with log⁡0.55\log_{0.5}5.

Since 2>12>1, log⁡2x\log_2 x is increasing:

log⁡23<log⁡25.\log_2 3<\log_2 5.

Since 0<0.5<10<0.5<1, log⁡0.5x\log_{0.5}x is decreasing:

log⁡0.53>log⁡0.55.\log_{0.5}3>\log_{0.5}5.

When comparing logarithms with the same valid base, check whether the function increases or decreases before choosing the inequality sign.

Example 2: Simplify a product

Evaluate

(log⁡43+log⁡83)(log⁡32+log⁡92).(\log_4 3+\log_8 3)(\log_3 2+\log_9 2).

Changing the bases to 22 and 33 gives

log⁡43+log⁡83=(12+13)log⁡23=56log⁡23,\log_4 3+\log_8 3 =\left(\frac12+\frac13\right)\log_2 3 =\frac56\log_2 3, log⁡32+log⁡92=(1+12)log⁡32=32log⁡32.\log_3 2+\log_9 2 =\left(1+\frac12\right)\log_3 2 =\frac32\log_3 2.

Because log⁡23  log⁡32=1\log_2 3\;\log_3 2=1, the product is

56×32=54.\frac56\times\frac32=\boxed{\frac54}.

Example 3: Find the domain when the base also varies

Find the domain of y=log⁡2x−1(3−x)y=\log_{2x-1}(3-x).

Apply all three restrictions:

2x−1>0,2x−1≠1,3−x>0.2x-1>0,\qquad 2x-1\ne1,\qquad 3-x>0.

These give x>12x>\frac12, x≠1x\ne1, and x<3x<3. Their intersection is

(12,1)∪(1,3).\boxed{\left(\frac12,1\right)\cup(1,3)}.

Checking only the argument would miss both restrictions on the base.

5. Before you finish

Check the base and argument in the original expression, not just in a simplified form. Keep any excluded values when you transform an equation. For inequalities with a fixed base between 00 and 11, remember that applying the logarithm reverses the order.

Further reading

Content reviewed: 9 September 2026.