04 Logarithmic Functions

A structured guide to logarithmic functions, their inverse relationship with exponential functions, key properties, equation-solving methods, inequalities, and applications to growth and decay.

Logarithms and inverse functions

A answers the question: what exponent on a given base produces a specified positive number? Its definition is

log⁡bx=y⟺by=x,\log_b x=y \quad \Longleftrightarrow \quad b^y=x,

with b>0b>0, b≠1b\ne1, and x>0x>0.

For example, log⁡28=3\log_2 8=3 because 23=82^3=8, log⁡10(1100)=−2\log_{10}\left(\frac{1}{100}\right)=-2 because 10−2=110010^{-2}=\frac{1}{100}, and log⁡51=0\log_5 1=0 because 50=15^0=1.

Inverse relationships

The f(x)=bxf(x)=b^x and the logarithmic function f−1(x)=log⁡bxf^{-1}(x)=\log_b x are inverses. Therefore,

log⁡b(bx)=x\log_b\left(b^x\right)=x

for every real number xx, and

blog⁡bx=xb^{\log_b x}=x

for x>0x>0. Their graphs are reflections across the line y=xy=x.

and graph behavior

For y=log⁡bxy=\log_b x, the is

(0,∞),(0,\infty),

and the range is

(−∞,∞).(-\infty,\infty).

The graph has a vertical asymptote at x=0x=0, crosses the xx-axis at (1,0)(1,0), and has no yy-intercept. If b>1b>1, the function is increasing; if 0<b<10<b<1, it is decreasing.

The is the with base ee:

ln⁡x=log⁡ex,\ln x=\log_e x,

where e≈2.71828e\approx2.71828. Thus, ln⁡(ex)=x\ln(e^x)=x and eln⁡x=xe^{\ln x}=x for x>0x>0.

Properties and change of base

rules are consequences of exponent rules and apply when the relevant arguments are positive.

Product, quotient, and power rules

The states

log⁡b(MN)=log⁡bM+log⁡bN.\log_b(MN)=\log_b M+\log_b N.

For example,

log⁡2(8⋅4)=log⁡28+log⁡24=3+2=5.\log_2(8\cdot4)=\log_2 8+\log_2 4=3+2=5.

The quotient property is

log⁡b(MN)=log⁡bM−log⁡bN,\log_b\left(\frac{M}{N}\right)=\log_b M-\log_b N,

and the power property is

log⁡b(Mr)=rlog⁡bM.\log_b(M^r)=r\log_b M.

For instance,

log⁡5(253)=3log⁡525=3(2)=6.\log_5(25^3)=3\log_5 25=3(2)=6.

The rules can be used in reverse. If the arguments are positive, then

2ln⁡x+ln⁡(x+1)=ln⁡(x2)+ln⁡(x+1)=ln⁡(x2(x+1)).2\ln x+\ln(x+1)=\ln(x^2)+\ln(x+1)=\ln\bigl(x^2(x+1)\bigr).

Nonproperties

Logarithms do not distribute over addition or subtraction. In general,

log⁡b(M+N)≠log⁡bM+log⁡bN\log_b(M+N)\ne\log_b M+\log_b N

and

log⁡b(M−N)≠log⁡bM−log⁡bN.\log_b(M-N)\ne\log_b M-\log_b N.

The product and quotient rules apply only to multiplication and division inside the .

Changing the base

The is

log⁡bM=log⁡Mlog⁡b=ln⁡Mln⁡b.\log_b M=\frac{\log M}{\log b}=\frac{\ln M}{\ln b}.

For example,

log⁡720=ln⁡20ln⁡7≈1.540.\log_7 20=\frac{\ln20}{\ln7}\approx1.540.

Takeaway: Products become sums, quotients become differences, and powers become coefficients; sums and differences inside logarithms cannot be split.

Solving logarithmic equations

Before solving, determine the : every argument must be greater than zero. This condition prevents invalid values involving the of zero or a negative number.

A single

For

log⁡b(S)=c,\log_b(S)=c,

rewrite in exponential form:

S=bc.S=b^c.

For example,

log⁡4(2x−1)=3\log_4(2x-1)=3

becomes

2x−1=43=64,2x-1=4^3=64,

so

x=652.x=\frac{65}{2}.

The original argument is positive for this value, so it is valid.

Equal logarithms

For logarithms with the same base,

log⁡bS=log⁡bT⟹S=T.\log_b S=\log_b T \quad \Longrightarrow \quad S=T.

Thus,

log⁡3(x+4)=log⁡3(2x−1)\log_3(x+4)=\log_3(2x-1)

gives

x+4=2x−1,x+4=2x-1,

so x=5x=5. Both arguments are positive at this value.

Combining logarithms

Consider

ln⁡x+ln⁡(x−3)=ln⁡4.\ln x+\ln(x-3)=\ln4.

The requires x>0x>0 and x−3>0x-3>0, hence x>3x>3. Combining the left side gives

ln⁡(x(x−3))=ln⁡4,\ln\bigl(x(x-3)\bigr)=\ln4,

so

x(x−3)=4.x(x-3)=4.

Factoring gives

(x−4)(x+1)=0.(x-4)(x+1)=0.

The candidates are x=4x=4 and x=−1x=-1, but only x=4x=4 satisfies x>3x>3. Always check candidates in the original equation.

Solving exponential equations

Logarithms are especially useful when the unknown occurs in an exponent.

Same base

If both sides can be written with the same base, equate exponents. For example,

2x+1=16=242^{x+1}=16=2^4

implies

x+1=4,x+1=4,

so x=3x=3.

Different bases

For

5x=17,5^x=17,

take natural logarithms:

ln⁡(5x)=ln⁡17.\ln(5^x)=\ln17.

Using the power property gives

xln⁡5=ln⁡17,x\ln5=\ln17,

and therefore

x=ln⁡17ln⁡5≈1.760.x=\frac{\ln17}{\ln5}\approx1.760.

Shifted exponential equations

First isolate the exponential expression. To solve

3e2x−5=10,3e^{2x}-5=10,

add 55 and divide by 33:

e2x=5.e^{2x}=5.

Taking the gives

2x=ln⁡5,2x=\ln5,

so

x=ln⁡52.x=\frac{\ln5}{2}.

Takeaway: Isolate the exponential term before taking logarithms, then use the power property to bring the exponent down.

Logarithmic inequalities

A is controlled by the base because the base determines whether the logarithmic function is increasing or decreasing.

Base greater than one

When b>1b>1, the function is increasing and the inequality direction is preserved:

log⁡bS<log⁡bT⟺S<T.\log_b S<\log_b T \quad \Longleftrightarrow \quad S<T.

For

log⁡2(x−1)≤3,\log_2(x-1)\le3,

conversion to exponential form gives

x−1≤23=8.x-1\le2^3=8.

The requires x−1>0x-1>0, or x>1x>1. Combining the conditions gives

1<x≤9.1<x\le9.

Base between zero and one

When 0<b<10<b<1, the function is decreasing and the inequality direction reverses. Consider

log⁡1/3(x+2)>2.\log_{1/3}(x+2)>2.

Conversion gives

x+2<(13)2=19.x+2<\left(\frac{1}{3}\right)^2=\frac{1}{9}.

The requires x+2>0x+2>0, or x>−2x>-2, while the converted inequality requires x<19−2=−179x<\frac{1}{9}-2=-\frac{17}{9}. These conditions are incompatible because −179<−2-\frac{17}{9}<-2, so there is no real solution.

Reliable procedure

  1. State the positivity condition for every argument.

  2. Determine whether the base is greater than 11 or between 00 and 11.

  3. Convert or compare expressions, reversing the inequality only when the base is between 00 and 11.

  4. Intersect the result with the .

Takeaway: Check both monotonicity and before accepting a solution.

Applications to growth, decay, and scale

Logarithms allow exponential models to be solved for time and help represent quantities spanning very large or very small ranges.

Continuous growth and decay

A continuous model has the form

A(t)=A0ekt,A(t)=A_0e^{kt},

where A0A_0 is the initial amount, kk is the growth or decay constant, and tt is time. Solving for tt gives

AA0=ekt,\frac{A}{A_0}=e^{kt},
ln⁡(AA0)=kt,\ln\left(\frac{A}{A_0}\right)=kt,

and therefore

t=1kln⁡(AA0).t=\frac{1}{k}\ln\left(\frac{A}{A_0}\right).

For decay written as A(t)=A0e−rtA(t)=A_0e^{-rt}, where r>0r>0,

t=1rln⁡(A0A).t=\frac{1}{r}\ln\left(\frac{A_0}{A}\right).

Example: continuous population growth

Suppose a population starts at 500500 and grows continuously at 4%4\% per year. Its model is

P(t)=500e0.04t.P(t)=500e^{0.04t}.

To find when it reaches 800800, solve

800=500e0.04t.800=500e^{0.04t}.

After division and taking natural logarithms,

ln⁡(1.6)=0.04t,\ln(1.6)=0.04t,

so

t=ln⁡(1.6)0.04≈11.75.t=\frac{\ln(1.6)}{0.04}\approx11.75.

The population reaches 800800 after approximately 11.7511.75 years.

For a decay model A(t)=A0ektA(t)=A_0e^{kt} with k<0k<0, the TT satisfies

12A0=A0ekT.\frac{1}{2}A_0=A_0e^{kT}.

After dividing by A0A_0 and taking natural logarithms,

ln⁡(12)=kT.\ln\left(\frac{1}{2}\right)=kT.

Since ln⁡(1/2)=−ln⁡2\ln(1/2)=-\ln2,

T=−ln⁡2k.T=-\frac{\ln2}{k}.

Logarithmic scales

A logarithmic scale represents multiplicative changes through additive distances. This makes logarithms useful for measurements such as sound intensity, acidity, and earthquake magnitude. The exact interpretation depends on the scale and its reference value.

Problem-solving checklist and summary

Use the following workflow to organize solutions:

  1. Identify the . Require every argument to be positive.

  2. Choose a useful form. Use the definition, exponential form, or properties as appropriate.

  3. Apply properties carefully. Product, quotient, and power rules do not apply to sums or differences inside a .

  4. For exponential equations, isolate the exponential expression before taking logarithms.

  5. For logarithmic equations, convert a single to exponential form or combine logarithms before comparing arguments.

  6. For inequalities, determine whether the base is greater than 11 or between 00 and 11.

  7. Check every candidate in the original equation or inequality.

  8. Round only at the end when using a calculator.

The central ideas are:

  • log⁡bx=y\log_b x=y means by=xb^y=x.

  • The base satisfies b>0b>0 and b≠1b\ne1.

  • Every argument must be positive.

  • Logarithms reverse exponential operations.

  • Products become sums, quotients become differences, and powers become coefficients.

  • The is ln⁡x=log⁡ex\ln x=\log_e x.

  • The is log⁡bM=ln⁡Mln⁡b\log_b M=\frac{\ln M}{\ln b}.

  • Inequality direction is preserved for bases greater than 11 and reversed for bases between 00 and 11.

  • Logarithms help solve for unknown exponents and model growth, decay, , and multiplicative scales.