Logarithmic functions are the inverse of exponential functions, allowing us to solve complex equations and model real-world phenomena. They're essential for understanding growth, decay, and scale in fields like physics, chemistry, and finance.
In this section, we'll learn how to convert between logarithmic and exponential forms, evaluate logarithmic expressions, and apply key properties. We'll also explore graphing techniques and practical applications in earthquake measurement, sound intensity, and radioactive decay.
Logarithmic Functions more resources to help you study Logarithmic and exponential conversions
Logarithmic form log b ( x ) = y \log_b(x) = y log b ( x ) = y equivalent to exponential form b y = x b^y = x b y = x
b b b represents the base, x x x the argument, and y y y the logarithm
Convert from logarithmic to exponential form using b log b ( x ) = x b^{\log_b(x)} = x b l o g b ( x ) = x
Example: 2 log 2 ( 8 ) = 8 2^{\log_2(8)} = 8 2 l o g 2 ( 8 ) = 8 because log 2 ( 8 ) = 3 \log_2(8) = 3 log 2 ( 8 ) = 3 and 2 3 = 8 2^3 = 8 2 3 = 8
Convert from exponential to logarithmic form using log b ( b x ) = x \log_b(b^x) = x log b ( b x ) = x
Example: log 5 ( 5 2 ) = 2 \log_5(5^2) = 2 log 5 ( 5 2 ) = 2 because 5 2 = 25 5^2 = 25 5 2 = 25 and log 5 ( 25 ) = 2 \log_5(25) = 2 log 5 ( 25 ) = 2
Logarithmic functions are inverse functions of exponential functions
Evaluation of logarithmic expressions
Common logarithm log 10 ( x ) \log_{10}(x) log 10 ( x ) often written as log ( x ) \log(x) log ( x )
log ( 1000 ) = 3 \log(1000) = 3 log ( 1000 ) = 3 because 10 3 = 1000 10^3 = 1000 1 0 3 = 1000
log ( 0.01 ) = − 2 \log(0.01) = -2 log ( 0.01 ) = − 2 because 10 − 2 = 0.01 10^{-2} = 0.01 1 0 − 2 = 0.01
Natural logarithm log e ( x ) \log_e(x) log e ( x ) often written as ln ( x ) \ln(x) ln ( x )
e e e mathematical constant approximately 2.71828
ln ( e 4 ) = 4 \ln(e^4) = 4 ln ( e 4 ) = 4 because e 4 = e 4 e^4 = e^4 e 4 = e 4
ln ( 1 ) = 0 \ln(1) = 0 ln ( 1 ) = 0 because e 0 = 1 e^0 = 1 e 0 = 1
Change of base formula log b ( x ) = log a ( x ) log a ( b ) \log_b(x) = \frac{\log_a(x)}{\log_a(b)} log b ( x ) = l o g a ( b ) l o g a ( x ) where a a a any positive base
Example: log 3 ( 9 ) = ln ( 9 ) ln ( 3 ) ≈ 2 \log_3(9) = \frac{\ln(9)}{\ln(3)} \approx 2 log 3 ( 9 ) = l n ( 3 ) l n ( 9 ) ≈ 2 because 3 2 = 9 3^2 = 9 3 2 = 9
Properties and applications of logarithms
Product rule log b ( M N ) = log b ( M ) + log b ( N ) \log_b(MN) = \log_b(M) + \log_b(N) log b ( MN ) = log b ( M ) + log b ( N )
log 2 ( 8 ⋅ 4 ) = log 2 ( 8 ) + log 2 ( 4 ) = 3 + 2 = 5 \log_2(8 \cdot 4) = \log_2(8) + \log_2(4) = 3 + 2 = 5 log 2 ( 8 ⋅ 4 ) = log 2 ( 8 ) + log 2 ( 4 ) = 3 + 2 = 5
Quotient rule log b ( M N ) = log b ( M ) − log b ( N ) \log_b(\frac{M}{N}) = \log_b(M) - \log_b(N) log b ( N M ) = log b ( M ) − log b ( N )
log 3 ( 81 9 ) = log 3 ( 81 ) − log 3 ( 9 ) = 4 − 2 = 2 \log_3(\frac{81}{9}) = \log_3(81) - \log_3(9) = 4 - 2 = 2 log 3 ( 9 81 ) = log 3 ( 81 ) − log 3 ( 9 ) = 4 − 2 = 2
Power rule log b ( M n ) = n log b ( M ) \log_b(M^n) = n \log_b(M) log b ( M n ) = n log b ( M )
log 2 ( 16 3 ) = 3 log 2 ( 16 ) = 3 ⋅ 4 = 12 \log_2(16^3) = 3 \log_2(16) = 3 \cdot 4 = 12 log 2 ( 1 6 3 ) = 3 log 2 ( 16 ) = 3 ⋅ 4 = 12
Zero exponent rule log b ( 1 ) = 0 \log_b(1) = 0 log b ( 1 ) = 0
log 5 ( 1 ) = 0 \log_5(1) = 0 log 5 ( 1 ) = 0 because 5 0 = 1 5^0 = 1 5 0 = 1
Identity rule log b ( b ) = 1 \log_b(b) = 1 log b ( b ) = 1
log 7 ( 7 ) = 1 \log_7(7) = 1 log 7 ( 7 ) = 1 because 7 1 = 7 7^1 = 7 7 1 = 7
Solving logarithmic equations
Isolate logarithm on one side of equation
Apply exponential function with same base to both sides
Solve resulting exponential equation
Example: Solve log 2 ( x ) = 3 \log_2(x) = 3 log 2 ( x ) = 3
log 2 ( x ) = 3 \log_2(x) = 3 log 2 ( x ) = 3
2 log 2 ( x ) = 2 3 2^{\log_2(x)} = 2^3 2 l o g 2 ( x ) = 2 3
x = 8 x = 8 x = 8
Graphing and analysis of logarithms
Parent function f ( x ) = log b ( x ) f(x) = \log_b(x) f ( x ) = log b ( x ) where b > 0 b > 0 b > 0 and b ≠ 1 b \neq 1 b = 1
Domain ( 0 , ∞ ) (0, \infty) ( 0 , ∞ ) , range ( − ∞ , ∞ ) (-\infty, \infty) ( − ∞ , ∞ ) , vertical asymptote x = 0 x = 0 x = 0
Increasing if b > 1 b > 1 b > 1 , decreasing if 0 < b < 1 0 < b < 1 0 < b < 1
Transformations
Vertical shift f ( x ) = log b ( x ) + k f(x) = \log_b(x) + k f ( x ) = log b ( x ) + k shifts graph up by k k k units
Horizontal shift f ( x ) = log b ( x − h ) f(x) = \log_b(x - h) f ( x ) = log b ( x − h ) shifts graph right by h h h units
Vertical stretch/compression f ( x ) = a log b ( x ) f(x) = a \log_b(x) f ( x ) = a log b ( x )
Stretches graph vertically by factor of ∣ a ∣ |a| ∣ a ∣ if ∣ a ∣ > 1 |a| > 1 ∣ a ∣ > 1
Compresses graph vertically if 0 < ∣ a ∣ < 1 0 < |a| < 1 0 < ∣ a ∣ < 1
Reflection f ( x ) = − log b ( x ) f(x) = -\log_b(x) f ( x ) = − log b ( x ) reflects graph over x x x -axis
Real-world logarithmic problem solving
Richter scale M = log ( I I 0 ) M = \log(\frac{I}{I_0}) M = log ( I 0 I )
M M M magnitude, I I I earthquake intensity, I 0 I_0 I 0 reference intensity
Example: Earthquake with intensity 1000 times reference has magnitude log ( 1000 ) = 3 \log(1000) = 3 log ( 1000 ) = 3
Decibel scale β = 10 log ( I I 0 ) \beta = 10 \log(\frac{I}{I_0}) β = 10 log ( I 0 I )
β \beta β sound intensity level in decibels, I I I sound intensity, I 0 I_0 I 0 reference intensity
Example: Sound 100 times reference intensity has decibel level 10 log ( 100 ) = 20 10 \log(100) = 20 10 log ( 100 ) = 20 dB
Exponential growth and decay A ( t ) = A 0 e k t A(t) = A_0 e^{kt} A ( t ) = A 0 e k t
A ( t ) A(t) A ( t ) amount at time t t t , A 0 A_0 A 0 initial amount, k k k growth or decay constant
Half-life t 1 / 2 = ln ( 0.5 ) k t_{1/2} = \frac{\ln(0.5)}{k} t 1/2 = k l n ( 0.5 ) where t 1 / 2 t_{1/2} t 1/2 half-life, k k k decay constant
Example: Radioactive substance half-life 10 days, decay constant k = ln ( 0.5 ) 10 ≈ − 0.0693 k = \frac{\ln(0.5)}{10} \approx -0.0693 k = 10 l n ( 0.5 ) ≈ − 0.0693
pH scale measures acidity using logarithms: p H = − log [ H + ] pH = -\log[H^+] p H = − log [ H + ]