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

Solution:

(a)

$
\log_2(8 \cdot 4) = \log_2 8 + \log_2 4 \quad \text{[Product Rule]}
$

$
= 3 + 2 = 5
$

(b)

$
\log_5\left(\frac{125}{25}\right) = \log_5 125 – \log_5 25 \quad \text{[Quotient Rule]}
$

$
= 3 – 2 = 1
$

(c)

$
\log_3(27^2) = 2 \cdot \log_3 27 \quad \text{[Power Rule]}
$

$
= 2 \cdot 3 = 6
$

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, \(x = \rm{3}\).

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:

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

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