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IB Mathematics AI SL Laws of logarithms Study Notes - New Syllabus

IB Mathematics AI SL Laws of logarithms Study Notes

LEARNING OBJECTIVE

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Key Concepts: 

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MAI HL and SL Notes – All topics

Logarithms

♦Definition

A logarithm is the power to which a number (called base) must be raised to obtain another number.

Notation: If \(a^{x} = b\), then \(\log_{a}b = x\).

♦Properties:

 \(\log_{a}1 = 0\)
 \(\log_{a}a = 1\)
 \(\log_{a}(x \cdot y) = \log_{a}x + \log_{a}y\)
 \(\log_{a}\left(\frac{x}{y}\right) = \log_{a}x – \log_{a}y\)
 \(\log_{a}x^{p} = p\log_{a}x\), where \(p\) is a constant
 \(\log_{a}b = \frac{1}{\log_{b}a}\) (change of base formula)

Example

If \(2^{x} = 8\), then

▶️Answer/Explanation

Solution:

\(\log_{2}8 = x\).

♦GDC TI-84 Plus: To evaluate logarithms using GDC TI-84 Plus, use the log button followed by the base, then the argument in parentheses. For example, to evaluate \(\log_{2}8\), enter \(\log(2,8)\) and press ENTER.

Note: When the base is not specified, the logarithm is assumed to be base 10.

 

Example

Solve for \(x\):

\(
\log_{2}(x-1) + \log_{2}(x+1) = 3
\)

▶️Answer/Explanation

Solution:

Using the logarithmic rule \(\log_{a}(m) + \log_{a}(n) = \log_{a}(mn)\), we can simplify the left side of the equation:

\(
\log_{2}[(x-1)(x+1)] = \log_{2}(x^{2}-1)
\)

So now we have:

\(
\log_{2}(x^{2}-1) = 3
\)

Using the exponential form of logarithms, we can write:

\(
2^{3} = x^{2}-1
\)

Solving for \(x\), we get:

\(
x = \pm\sqrt{9}
\)

However, we need to discard the negative solution since \(\log_{2}(x-1)\) and \(\log_{2}(x+1)\) are only defined for positive values of \(x\). Therefore, \(x = \rm{3}\).

 

Example

Solve for \(x\):

\(
\ln(2x+1) = 3 + \ln(5x-1)
\)

▶️Answer/Explanation

Solution:

Using the logarithmic rule \(\ln(m) + \ln(n) = \ln(mn)\), we can simplify the right side of the equation:

\(
\ln[(5x-1)e^{3}] = \ln(5x-1) + 3\ln(e) = \ln(5x-1) + 3
\)

So now we have:

\(
\ln(2x+1) = \ln(5x-1) + 3
\)

Using the logarithmic rule \(\ln(m) – \ln(n) = \ln\left(\frac{m}{n}\right)\), we can further simplify the equation:

\(
\ln\left(\frac{2x+1}{5x-1}\right) = 3
\)

Using the exponential form of logarithms, we can write:

\(
e^{3} = \frac{2x+1}{5x-1}
\)

Solving for \(x\), we get:

\(
x = \frac{e^{3} + 1}{5e^{3} – 2}
\)

Calculating the numerical value (approximating \(e^{3} \approx 20.0855\)):

\(
x \approx \frac{20.0855 + 1}{5 \times 20.0855 – 2} = \frac{21.0855}{98.4275} \approx 0.2142
\)

Verification:
Ensure the arguments of the logarithms are positive:
\(2x + 1 > 0 \implies x > -0.5\)
 \(5x – 1 > 0 \implies x > 0.2\)

Since \(0.2142 > 0.2\), the solution is valid.

Therefore, the exact solution is:

\(
x = \frac{e^{3} + 1}{5e^{3} – 2}
\)

\(
x \approx \rm{0.2142}
\)

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