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Functions and Their Modeling

Logarithmic Function Identity

Understanding Logarithmic Identities

Logarithmic identities are special properties that apply to all logarithmic functions. These properties are very helpful in simplifying calculations and solving complex logarithmic equations.

Before discussing logarithmic identities, let's recall that logarithms are the inverse of exponents. If bx=ab^x = a, then blog⁡a=x^b\log a = x.

Basic Logarithmic Identities

Product Identity

blog⁡(MN)=blog⁡M+blog⁡N^b\log(MN) = {^b\log M} + {^b\log N}

The logarithm of a product equals the sum of the logarithms of each number.

Example:

2log⁡(8×4)=2log⁡8+2log⁡4=3+2=5^2\log(8 \times 4) = {^2\log 8} + {^2\log 4} = 3 + 2 = 5

Quotient Identity

blog⁡(MN)=blog⁡M−blog⁡N^b\log\left(\frac{M}{N}\right) = {^b\log M} - {^b\log N}

The logarithm of a quotient equals the difference between the logarithm of the numerator and the logarithm of the denominator.

Example:

3log⁡(819)=3log⁡81−3log⁡9=4−2=2^3\log\left(\frac{81}{9}\right) = {^3\log 81} - {^3\log 9} = 4 - 2 = 2

Power Identity

blog⁡(Mp)=p⋅blog⁡M^b\log(M^p) = p \cdot {^b\log M}

The logarithm of a number raised to a power equals the power multiplied by the logarithm of that number.

Example:

2log⁡(43)=3⋅2log⁡4=3×2=6^2\log(4^3) = 3 \cdot {^2\log 4} = 3 \times 2 = 6

Special Logarithmic Identities

Change of Base

blog⁡M=alog⁡Malog⁡b^b\log M = \frac{^a\log M}{^a\log b}

This identity allows us to change the logarithm base as needed.

Example:

3log⁡9=10log⁡910log⁡3=0.9540.477=2^3\log 9 = \frac{^{10}\log 9}{^{10}\log 3} = \frac{0.954}{0.477} = 2

Equality Identity

If blog⁡M=blog⁡N^b\log M = {^b\log N}, then M=NM = N

Two numbers that have the same logarithmic value (with the same base) must be the same number.

Inequality Identity

  • If b>1b > 1 and blog⁡M<blog⁡N^b\log M < {^b\log N}, then M<NM < N
  • If 0<b<10 < b < 1 and blog⁡M<blog⁡N^b\log M < {^b\log N}, then M>NM > N

Applications of Logarithmic Identities

Simplifying Expressions

Simplify 2log⁡32+2log⁡8−2log⁡4^2\log 32 + {^2\log 8} - {^2\log 4}

Solution:

2log⁡32+2log⁡8−2log⁡4^2\log 32 + {^2\log 8} - {^2\log 4}
=2log⁡(32×8)−2log⁡4= {^2\log(32 \times 8)} - {^2\log 4}
=2log⁡256−2log⁡4= {^2\log 256} - {^2\log 4}
=2log⁡(2564)= {^2\log\left(\frac{256}{4}\right)}
=2log⁡64=6= {^2\log 64} = 6

Solving Equations

Find the value of xx if 3log⁡x+3log⁡9=3log⁡81^3\log x + {^3\log 9} = {^3\log 81}

Solution:

3log⁡x+3log⁡9=3log⁡81^3\log x + {^3\log 9} = {^3\log 81}
3log⁡(x×9)=3log⁡81^3\log(x \times 9) = {^3\log 81}
x×9=81x \times 9 = 81
x=819=9x = \frac{81}{9} = 9

Real-Life Applications

Richter Scale

Earthquake strength is measured using the Richter scale which is based on logarithms:

R=log⁡(II0)R = \log\left(\frac{I}{I_0}\right)

Where:

  • RR = Richter scale value
  • II = earthquake intensity
  • I0I_0 = reference intensity (zero level)

Example: An earthquake that occurred in Haiti in 2010 had an intensity of 10710^7 times compared to zero-level earthquakes. What is the Richter scale strength of that earthquake?

Solution:

R=log⁡(II0)R = \log\left(\frac{I}{I_0}\right)
=log⁡(107I0I0)= \log\left(\frac{10^7 I_0}{I_0}\right)
=log⁡(107)= \log(10^7)
=7= 7

Therefore, the earthquake in Haiti in 2010 had a strength of 7 on the Richter scale.

Battery Charging

Battery charging time can be calculated using the logarithmic formula:

t=−1kln⁡(1−CC0)t = -\frac{1}{k}\ln\left(1 - \frac{C}{C_0}\right)

Where:

  • tt = charging time (in minutes)
  • kk = charging constant
  • CC = desired capacity
  • C0C_0 = maximum capacity

Example: Determine the time required to charge a battery from empty to 90% full. Assume k=0.02k = 0.02.

Solution:

t=−1kln⁡(1−CC0)t = -\frac{1}{k}\ln\left(1 - \frac{C}{C_0}\right)
=−10.02ln⁡(1−0.9C0C0)= -\frac{1}{0.02}\ln\left(1 - \frac{0.9C_0}{C_0}\right)
=−50ln⁡(1−0.9)= -50\ln(1 - 0.9)
=−50ln⁡(0.1)= -50\ln(0.1)
≈115.13\approx 115.13

Therefore, the charging time is approximately 115 minutes.

Car Price Depreciation

Logarithmic functions are also used for modeling decay/depreciation with the formula:

H(t)=cektH(t) = ce^{kt}

where H(t)H(t) is the value at time tt.

Example: At any given time, the price of a used car is not proportional to its current price. If a new car costs 200 million rupiah and after 5 years becomes 100 million rupiah, determine the car's price after 10 years of use.

Solution:

H(0)=200 million, so 200=ce0=cH(0) = 200 \text{ million, so } 200 = ce^0 = c
H(5)=100 million, so 100=200e5kH(5) = 100 \text{ million, so } 100 = 200e^{5k}
e5k=12, therefore 5k=ln⁡(12)e^{5k} = \frac{1}{2} \text{, therefore } 5k = \ln\left(\frac{1}{2}\right)
k=15ln⁡(12)=−0.1386k = \frac{1}{5}\ln\left(\frac{1}{2}\right) = -0.1386

From these results, the car's price at any time tt is:

H(t)=200e−0.1386tH(t) = 200e^{-0.1386t}

Therefore, the car's price after 10 years of use is:

H(10)=200e−0.1386(10)=200e−1.386≈50 million rupiahH(10) = 200e^{-0.1386(10)} = 200e^{-1.386} \approx 50 \text{ million rupiah}

Exercises

Problem 1

Simplify: 5log⁡125+5log⁡25−5log⁡5^5\log 125 + {^5\log 25} - {^5\log 5}

Problem 2

If 2log⁡x=3^2\log x = 3 and 2log⁡y=5^2\log y = 5, find the value of 2log⁡(xy)^2\log(xy)

Problem 3

Find the value of xx if 4log⁡(x−1)=4log⁡16−4log⁡2^4\log(x-1) = {^4\log 16} - {^4\log 2}

Answer Key

Answer 1

5log⁡125+5log⁡25−5log⁡5^5\log 125 + {^5\log 25} - {^5\log 5}
=3+2−1=4= 3 + 2 - 1 = 4

Answer 2

2log⁡(xy)=2log⁡x+2log⁡y^2\log(xy) = {^2\log x} + {^2\log y}
=3+5=8= 3 + 5 = 8

Answer 3

4log⁡(x−1)=4log⁡16−4log⁡2^4\log(x-1) = {^4\log 16} - {^4\log 2}
4log⁡(x−1)=4log⁡(162)^4\log(x-1) = {^4\log\left(\frac{16}{2}\right)}
4log⁡(x−1)=4log⁡8^4\log(x-1) = {^4\log 8}
x−1=8x - 1 = 8
x=9x = 9
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