The Central Idea
Logarithms return exponents.
Because exponents turn multiplication of same-base powers into addition of exponents, logarithms turn multiplication into addition:
\[ \boxed{\log_b(MN)=\log_b(M)+\log_b(N)}. \]Similarly, division becomes subtraction and powers become multiplication.
These are structural consequences of exponent laws, not unrelated formulas invented for logarithms.
Conditions on Logarithm Laws
Throughout the real-number logarithm laws, the base must satisfy
\[ \boxed{b>0,\qquad b\ne1}. \]Each logarithmic argument must also be positive.
Thus expressions such as
\[ \log_b(M) \]require
\[ \boxed{M>0}. \]The Product Law
For positive \(M\) and \(N\),
\[ \boxed{ \log_b(MN) = \log_b(M)+\log_b(N) }. \]A logarithm of a product becomes a sum of logarithms.
Why the Product Law Works
Let
\[ x=\log_b(M) \]and
\[ y=\log_b(N). \]Then
\[ b^x=M \]and
\[ b^y=N. \]Multiply:
\[ MN=b^xb^y. \]The exponent product law gives
\[ MN=b^{x+y}. \]Therefore,
\[ \log_b(MN)=x+y. \]Substituting the original logarithms,
\[ \boxed{ \log_b(MN)=\log_b(M)+\log_b(N) }. \]Example: Expanding a Product
Expand
\[ \log_3(9x), \]assuming \(x>0\).
Use the product law:
\[ \log_3(9x) = \log_3(9)+\log_3(x). \]Since
\[ \log_3(9)=2, \]we obtain
\[ \boxed{2+\log_3(x)}. \]Example: Condensing a Sum
Condense
\[ \log_5(x)+\log_5(y), \]where \(x>0\) and \(y>0\).
Reverse the product law:
\[ \boxed{\log_5(xy)}. \]The Quotient Law
For positive \(M\) and \(N\),
\[ \boxed{ \log_b\left(\frac{M}{N}\right) = \log_b(M)-\log_b(N) }. \]A logarithm of a quotient becomes a difference of logarithms.
Why the Quotient Law Works
Let
\[ x=\log_b(M) \]and
\[ y=\log_b(N). \]Then
\[ M=b^x,\qquad N=b^y. \]Divide:
\[ \frac{M}{N} = \frac{b^x}{b^y}. \]The exponent quotient law gives
\[ \frac{M}{N}=b^{x-y}. \]Therefore,
\[ \boxed{ \log_b\left(\frac{M}{N}\right) = \log_b(M)-\log_b(N) }. \]Example: Expanding a Quotient
Expand
\[ \log_2\left(\frac{x}{8}\right), \]with \(x>0\).
Use the quotient law:
\[ \log_2(x)-\log_2(8). \]Since
\[ \log_2(8)=3, \]the result is
\[ \boxed{\log_2(x)-3}. \]Example: Condensing a Difference
Condense
\[ \log_7(x)-\log_7(y), \]where \(x>0\) and \(y>0\).
Reverse the quotient law:
\[ \boxed{\log_7\left(\frac{x}{y}\right)}. \]The Power Law
For a positive logarithmic argument,
\[ \boxed{ \log_b(M^p) = p\log_b(M) }. \]An exponent inside a logarithm becomes a multiplier outside.
Why the Power Law Works
Let
\[ x=\log_b(M). \]Then
\[ M=b^x. \]Raise both sides to power \(p\):
\[ M^p=(b^x)^p. \]The power-of-a-power law gives
\[ M^p=b^{px}. \]Therefore,
\[ \log_b(M^p)=px. \]Since \(x=\log_b(M)\),
\[ \boxed{ \log_b(M^p)=p\log_b(M) }. \]Example: Expanding a Power
Expand
\[ \log_4(x^5), \]under conditions making the original logarithm real and the stated expansion valid.
The power law gives
\[ \boxed{5\log_4(x)} \]when \(x>0\).
Example: Condensing a Coefficient
Condense
\[ 3\log_2(x), \]with \(x>0\).
Move the coefficient into the logarithm as an exponent:
\[ \boxed{\log_2(x^3)}. \]A Compact Logarithm-Law Reference
\[ \boxed{ \log_b(MN) = \log_b(M)+\log_b(N) } \] \[ \boxed{ \log_b\left(\frac{M}{N}\right) = \log_b(M)-\log_b(N) } \] \[ \boxed{ \log_b(M^p) = p\log_b(M) } \]These laws assume valid logarithmic bases and positive real logarithmic arguments wherever the individual logarithms appear.
Multiplication Becomes Addition
The product law is worth interpreting conceptually:
\[ \log_b(MN) = \log_b(M)+\log_b(N). \]Multiplication inside the logarithm becomes addition outside.
This is the inverse-side reflection of
\[ b^xb^y=b^{x+y}. \]Division Becomes Subtraction
The quotient law says
\[ \log_b\left(\frac{M}{N}\right) = \log_b(M)-\log_b(N). \]Division inside the logarithm becomes subtraction outside.
This mirrors
\[ \frac{b^x}{b^y}=b^{x-y}. \]Powers Become Multipliers
The power law says
\[ \log_b(M^p) = p\log_b(M). \]An exponent inside becomes an ordinary multiplier outside.
This mirrors
\[ (b^x)^p=b^{px}. \]Expanding Logarithms
To expand a logarithmic expression means to use logarithm laws to separate products, quotients, and powers into simpler logarithmic pieces.
For example,
\[ \log_b\left(\frac{x^3y}{z^2}\right) \]expands into
\[ \boxed{ 3\log_b(x)+\log_b(y)-2\log_b(z) }, \]assuming \(x>0\), \(y>0\), and \(z>0\).
Condensing Logarithms
To condense means to reverse the logarithm laws and combine several logarithms into one.
For example,
\[ 2\log_b(x)+\log_b(y)-\log_b(z) \]becomes
\[ \boxed{ \log_b\left(\frac{x^2y}{z}\right) }. \]Example: Full Expansion
Expand
\[ \log_3\left(\frac{9x^4\sqrt{y}}{z^2}\right), \]assuming \(x>0\), \(y>0\), and \(z>0\).
Start with the quotient law:
\[ \log_3(9x^4\sqrt y)-\log_3(z^2). \]Expand the numerator product:
\[ \log_3(9)+\log_3(x^4)+\log_3(y^{1/2}) - \log_3(z^2). \]Apply the power law:
\[ 2+4\log_3(x)+\frac12\log_3(y)-2\log_3(z). \]Therefore,
\[ \boxed{ 2+4\log_3(x)+\frac12\log_3(y)-2\log_3(z) }. \]Example: Full Condensation
Condense
\[ 2\log_5(x)+3\log_5(y)-\frac12\log_5(z), \]assuming positive variables.
Move coefficients into exponents:
\[ \log_5(x^2)+\log_5(y^3)-\log_5(z^{1/2}). \]Combine the sum:
\[ \log_5(x^2y^3)-\log_5(z^{1/2}). \]Combine the difference:
\[ \boxed{ \log_5\left(\frac{x^2y^3}{\sqrt z}\right) }. \]Known Logarithmic Values Can Simplify Expressions
Consider
\[ \log_2(32x). \]Expand:
\[ \log_2(32)+\log_2(x). \]Since
\[ \log_2(32)=5, \]we obtain
\[ \boxed{5+\log_2(x)}. \]Roots Can Be Handled with the Power Law
Since
\[ \sqrt[n]{x}=x^{1/n}, \]logarithms of roots can be rewritten as
\[ \log_b(\sqrt[n]{x}) = \log_b(x^{1/n}). \]Therefore,
\[ \boxed{ \log_b(\sqrt[n]{x}) = \frac1n\log_b(x) }, \]under the appropriate real-domain conditions.
Example: Logarithm of a Square Root
Simplify
\[ \log_7(\sqrt{x}), \]with \(x>0\).
Rewrite the radical:
\[ \sqrt{x}=x^{1/2}. \]Apply the power law:
\[ \boxed{ \frac12\log_7(x) }. \]Reciprocals Produce Negative Logarithms
For \(x>0\),
\[ \log_b\left(\frac1x\right) = \log_b(1)-\log_b(x). \]Since
\[ \log_b(1)=0, \]we obtain
\[ \boxed{ \log_b\left(\frac1x\right) = -\log_b(x) }. \]Logarithms of Powers of the Base
The inverse identity gives
\[ \log_b(b^x)=x. \]The power law reaches the same result:
\[ \log_b(b^x) = x\log_b(b). \]Since
\[ \log_b(b)=1, \]we again obtain
\[ \boxed{x}. \]The Product Law Does Not Apply to Addition
There is no logarithm law that says
\[ \log_b(M+N) = \log_b(M)+\log_b(N). \]In general,
\[ \boxed{ \log_b(M+N) \ne \log_b(M)+\log_b(N) }. \]Numerical Check of the Addition Error
Consider base \(10\):
\[ \log(10+100) = \log(110). \]But
\[ \log(10)+\log(100) = 1+2 = 3. \]Since
\[ \log(110)\ne3, \]logarithms do not distribute across addition.
The Quotient Law Does Not Apply to Subtraction
Similarly,
\[ \log_b(M-N) \]cannot generally be rewritten as
\[ \log_b(M)-\log_b(N). \]The quotient law applies to
\[ \log_b\left(\frac{M}{N}\right), \]not to a difference \(M-N\).
Domain Restrictions Can Be Hidden by Expansion
Consider
\[ \log_b(x^2). \]The original logarithm is defined whenever
\[ x^2>0, \]which means
\[ x\ne0. \]Writing
\[ 2\log_b(x) \]would require \(x>0\), so it is not equivalent over the entire real domain of the original expression.
The Absolute-Value Refinement
For real \(x\ne0\),
\[ x^2=|x|^2. \]Therefore,
\[ \log_b(x^2) = \log_b(|x|^2) = 2\log_b|x|. \]Thus the domain-preserving identity is
\[ \boxed{ \log_b(x^2)=2\log_b|x| }, \qquad x\ne0. \]Why Textbooks Often Assume Positive Variables
Many introductory logarithm exercises state that all variables are positive.
Under that assumption,
\[ |x|=x, \]and expansions such as
\[ \log_b(x^2)=2\log_b(x) \]can be used without repeatedly introducing absolute values.
The assumption simplifies notation, but it should not be forgotten.
Changing Bases
A logarithm may use any valid base, but calculators usually provide direct keys for only a small number of bases.
The change-of-base formula rewrites a logarithm in one base using logarithms in another base:
\[ \boxed{ \log_b(x) = \frac{\log_c(x)}{\log_c(b)} }. \]Conditions for Change of Base
In
\[ \log_b(x) = \frac{\log_c(x)}{\log_c(b)}, \]the original base must satisfy
\[ b>0,\qquad b\ne1, \]the new base must satisfy
\[ c>0,\qquad c\ne1, \]and the argument must satisfy
\[ x>0. \]Why the Change-of-Base Formula Works
Let
\[ y=\log_b(x). \]Then
\[ b^y=x. \]Take logarithm base \(c\) of both sides:
\[ \log_c(b^y)=\log_c(x). \]Apply the power law:
\[ y\log_c(b)=\log_c(x). \]Divide:
\[ y= \frac{\log_c(x)}{\log_c(b)}. \]Since \(y=\log_b(x)\),
\[ \boxed{ \log_b(x) = \frac{\log_c(x)}{\log_c(b)} }. \]Changing to Common Logarithms
Choosing base \(10\) gives
\[ \boxed{ \log_b(x) = \frac{\log(x)}{\log(b)} }. \]Here \(\log\) denotes the common logarithm, base \(10\).
Changing to Natural Logarithms
Choosing base \(e\) gives
\[ \boxed{ \log_b(x) = \frac{\ln(x)}{\ln(b)} }. \]Here \(\ln\) denotes the natural logarithm, base \(e\).
Either Common or Natural Logarithms Work
Both
\[ \frac{\log(x)}{\log(b)} \]and
\[ \frac{\ln(x)}{\ln(b)} \]equal
\[ \log_b(x). \]The numerator and denominator must simply use the same new base.
Example: Evaluate \(\log_2(10)\)
A calculator may not have a dedicated base-\(2\) logarithm key.
Use change of base:
\[ \log_2(10) = \frac{\ln(10)}{\ln(2)}. \]Numerically,
\[ \boxed{\log_2(10)\approx3.3219}. \]Example: Evaluate \(\log_7(20)\)
Use natural logarithms:
\[ \log_7(20) = \frac{\ln(20)}{\ln(7)}. \]Therefore,
\[ \boxed{\log_7(20)\approx1.5395}. \]Example: Change Base Using Common Logs
Evaluate
\[ \log_4(30). \]Using common logarithms,
\[ \log_4(30) = \frac{\log(30)}{\log(4)}. \]Therefore,
\[ \boxed{\log_4(30)\approx2.4534}. \]Change of Base Explains Reciprocal Logarithms
Consider
\[ \log_b(c). \]Change to base \(c\):
\[ \log_b(c) = \frac{\log_c(c)}{\log_c(b)}. \]Since
\[ \log_c(c)=1, \]we obtain
\[ \boxed{ \log_b(c)=\frac{1}{\log_c(b)} }. \]A Useful Reciprocal Identity
The previous result can be written
\[ \boxed{ \log_b(c)\log_c(b)=1 }. \]Equivalently,
\[ \boxed{ \log_b(c)=\frac1{\log_c(b)} }. \]Changing Bases Does Not Change the Logarithmic Value
The expression
\[ \log_b(x) \]represents one specific exponent.
Change of base does not alter that exponent. It merely computes it using a different logarithmic measuring system.
Why Change of Base Is Useful
Change of base is useful because it lets us:
- evaluate logarithms in arbitrary bases with a calculator;
- compare logarithmic expressions written with different bases;
- solve exponential equations numerically;
- rewrite formulas using the natural logarithm;
- derive relationships between different exponential scales.
Solving \(b^x=y\) with Change of Base
If
\[ b^x=y, \]then
\[ x=\log_b(y). \]Change base:
\[ \boxed{ x= \frac{\ln(y)}{\ln(b)} }. \]This is one of the most important practical uses of logarithms.
Example: Solve an Exponential Equation
Solve
\[ 3^x=20. \]Write
\[ x=\log_3(20). \]Change base:
\[ x= \frac{\ln20}{\ln3}. \]Therefore,
\[ \boxed{x\approx2.7268}. \]Logarithm Laws Can Expose Unknown Exponents
Consider
\[ A=Pb^t. \]Divide by \(P\):
\[ \frac{A}{P}=b^t. \]Take logarithms:
\[ \ln\left(\frac{A}{P}\right) = \ln(b^t). \]Apply the power law:
\[ \ln\left(\frac{A}{P}\right) = t\ln b. \]Therefore,
\[ \boxed{ t= \frac{\ln(A/P)}{\ln b} }. \]Logarithm Laws and Continuous Exponentials
Suppose
\[ A=A_0e^{kt}. \]Divide:
\[ \frac{A}{A_0}=e^{kt}. \]Apply natural logarithms:
\[ \ln\left(\frac{A}{A_0}\right)=kt. \]Therefore,
\[ \boxed{ t= \frac{1}{k} \ln\left(\frac{A}{A_0}\right) }. \]Common Error: Splitting a Sum
It is incorrect to write
\[ \log_b(x+y) = \log_b(x)+\log_b(y). \]The product law applies only when the argument is a product:
\[ \boxed{ \log_b(xy)=\log_b(x)+\log_b(y) }. \]Common Error: Splitting a Difference
It is also incorrect to write
\[ \log_b(x-y) = \log_b(x)-\log_b(y). \]The quotient law requires division:
\[ \boxed{ \log_b\left(\frac{x}{y}\right) = \log_b(x)-\log_b(y) }. \]Common Error: Moving a Coefficient Inside Incorrectly
If
\[ 3\log_b(x), \]is condensed, the \(3\) becomes an exponent:
\[ \boxed{ 3\log_b(x)=\log_b(x^3) }. \]It does not become
\[ \log_b(3x). \]Common Error: Dividing Logarithms in Change of Base Backward
The correct formula is
\[ \boxed{ \log_b(x) = \frac{\ln(x)}{\ln(b)} }. \]The logarithm of the argument goes in the numerator, and the logarithm of the original base goes in the denominator.
Common Error: Mixing Logarithm Types in One Ratio
Change of base requires the same logarithmic base in numerator and denominator.
Thus
\[ \frac{\ln(x)}{\ln(b)} \]is valid, and
\[ \frac{\log(x)}{\log(b)} \]is valid.
A mixed expression such as
\[ \frac{\ln(x)}{\log(b)} \]is not the standard change-of-base formula.
Common Error: Forgetting Positive Arguments
The expression
\[ \log_b(x)+\log_b(y) \]requires
\[ x>0,\qquad y>0. \]Condensing it to
\[ \log_b(xy) \]does not erase those original restrictions.
Common Error: Expanding \(\log_b(x^2)\) without Checking Domain
The original expression
\[ \log_b(x^2) \]allows any real
\[ x\ne0. \]The expression
\[ 2\log_b(x) \]allows only \(x>0\).
The domain-preserving real identity is
\[ \boxed{ \log_b(x^2)=2\log_b|x| }. \]A Reliable Procedure for Expanding Logarithms
- Record any stated domain assumptions.
- Use the quotient law to separate numerator and denominator.
- Use the product law to separate factors.
- Rewrite radicals as rational exponents when useful.
- Use the power law to move exponents in front.
- Evaluate any exact logarithms that are immediately recognizable.
- Check that the expanded form preserves the intended domain.
A Reliable Procedure for Condensing Logarithms
- Verify that the logarithms use the same base.
- Move coefficients into logarithms as exponents.
- Combine sums using the product law.
- Combine differences using the quotient law.
- Write the final result as one logarithm.
- Preserve restrictions from the original logarithmic arguments.
A Reliable Procedure for Changing Bases
- Identify the original base \(b\).
- Identify the argument \(x\).
- Choose a convenient new base, usually \(10\) or \(e\).
- Put the logarithm of the argument in the numerator.
- Put the logarithm of the original base in the denominator.
- Use the same new base in numerator and denominator.
- Evaluate and round only at the end if a decimal approximation is needed.
The pattern is
\[ \boxed{ \log_b(x) = \frac{\ln x}{\ln b} = \frac{\log x}{\log b} }. \]Example: Full Expansion Workflow
Expand
\[ \log_2\left(\frac{32x^3y^2}{\sqrt z}\right), \]assuming \(x>0\), \(y>0\), and \(z>0\).
Separate the quotient:
\[ \log_2(32x^3y^2)-\log_2(z^{1/2}). \]Separate the numerator factors:
\[ \log_2(32)+\log_2(x^3)+\log_2(y^2) - \log_2(z^{1/2}). \]Apply the power law:
\[ 5+3\log_2(x)+2\log_2(y)-\frac12\log_2(z). \]Therefore,
\[ \boxed{ 5+3\log_2(x)+2\log_2(y)-\frac12\log_2(z) }. \]Example: Full Condensing Workflow
Condense
\[ 4\ln x+2\ln y-\ln z, \]assuming positive variables.
Move coefficients into exponents:
\[ \ln(x^4)+\ln(y^2)-\ln z. \]Combine the sum:
\[ \ln(x^4y^2)-\ln z. \]Combine the quotient:
\[ \boxed{ \ln\left(\frac{x^4y^2}{z}\right) }. \]Example: Full Change-of-Base Workflow
Evaluate
\[ \log_6(17). \]Choose natural logarithms:
\[ \log_6(17) = \frac{\ln17}{\ln6}. \]Numerically,
\[ \boxed{ \log_6(17)\approx1.5812 }. \]Practice 1: Product Law
Expand
\[ \log_3(27x), \]assuming \(x>0\).
Practice 2: Quotient Law
Expand
\[ \log_5\left(\frac{x}{25}\right), \]assuming \(x>0\).
Practice 3: Power Law
Expand
\[ \ln(x^7), \]assuming \(x>0\).
Practice 4: Expand Completely
Expand
\[ \log_2\left(\frac{x^3y}{z^4}\right), \]assuming all variables are positive.
Practice 5: Condense
Condense
\[ \log_4(x)+2\log_4(y). \]Practice 6: Condense a Difference
Condense
\[ 3\ln x-\ln y. \]Practice 7: Change Base
Rewrite
\[ \log_7(15) \]using natural logarithms.
Practice 8: Approximate with Change of Base
Approximate
\[ \log_3(14). \]Practice 9: Solve an Exponential Equation
Solve
\[ 5^x=18 \]using logarithms.
Practice 10: Identify the Error
A student writes
\[ \log(x+4)=\log x+\log4. \]Explain why the step is invalid.
Practice Answers
Practice 1
\[ \log_3(27x) = \log_3(27)+\log_3(x). \]Since \(\log_3(27)=3\),
\[ \boxed{3+\log_3(x)}. \]Practice 2
\[ \log_5\left(\frac{x}{25}\right) = \log_5(x)-\log_5(25). \]Since \(\log_5(25)=2\),
\[ \boxed{\log_5(x)-2}. \]Practice 3
\[ \boxed{ \ln(x^7)=7\ln x }. \]Practice 4
\[ \log_2\left(\frac{x^3y}{z^4}\right) = \log_2(x^3)+\log_2(y)-\log_2(z^4). \]Therefore,
\[ \boxed{ 3\log_2(x)+\log_2(y)-4\log_2(z) }. \]Practice 5
\[ 2\log_4(y)=\log_4(y^2). \]Therefore,
\[ \boxed{ \log_4(xy^2) }. \]Practice 6
\[ 3\ln x=\ln(x^3). \]Therefore,
\[ \boxed{ \ln\left(\frac{x^3}{y}\right) }. \]Practice 7
\[ \boxed{ \log_7(15) = \frac{\ln15}{\ln7} }. \]Practice 8
\[ \log_3(14) = \frac{\ln14}{\ln3}. \]Therefore,
\[ \boxed{ \log_3(14)\approx2.4022 }. \]Practice 9
\[ 5^x=18. \]Take natural logarithms:
\[ \ln(5^x)=\ln18. \]Apply the power law:
\[ x\ln5=\ln18. \]Therefore,
\[ \boxed{ x=\frac{\ln18}{\ln5}\approx1.7959 }. \]Practice 10
The product law applies to multiplication:
\[ \log(x\cdot4)=\log x+\log4. \]It does not apply to addition.
Therefore,
\[ \boxed{ \log(x+4)\ne\log x+\log4 } \]in general.
What to Remember
The three central logarithm laws are
\[ \boxed{ \log_b(MN)=\log_b(M)+\log_b(N) } \] \[ \boxed{ \log_b\left(\frac{M}{N}\right) = \log_b(M)-\log_b(N) } \] \[ \boxed{ \log_b(M^p)=p\log_b(M) }. \]Change of base is
\[ \boxed{ \log_b(x) = \frac{\ln x}{\ln b} = \frac{\log x}{\log b} }. \]- Products inside logarithms become sums outside.
- Quotients inside logarithms become differences outside.
- Powers inside logarithms become multipliers outside.
- There is no corresponding rule for splitting sums or differences inside logarithms.
- Logarithmic arguments must remain positive in the real numbers.
- Expansion and condensation must preserve the domain of the original expression.
- Change of base lets any valid logarithm be evaluated using a convenient base such as \(10\) or \(e\).
The central habit is: look first at the operation inside the logarithm—product, quotient, or power—and apply only the law that matches that structure.
Where We Go Next
Logarithm laws give us the algebraic tools needed to expose unknown exponents, combine logarithmic expressions, and move between bases.
The next lesson, Solving Exponential and Logarithmic Equations, uses these tools systematically to isolate variables, solve equations exactly or approximately, identify extraneous solutions, and connect algebraic solutions with exponential and logarithmic graphs.