Logarithms (HSC SSCE Mathematics Advanced): Revision Notes

Worked Example 1: Converting from exponential to logarithmic form

Question: Rewrite $16^2 = 256$ in logarithmic form.

Strategy

We use the equivalence: if $a^y = x$, then $\log_a(x) = y$.

Working

Start with the exponential form:

$a^y = x$

Write the specific equation:

$16^2 = 256$

This tells us that base $a = 16$, argument $x = 256$, and exponent $y = 2$.

Now convert to logarithmic form:

$\log_a(x) = y$

Substitute our values:

$\log_{16}(256) = 2$

Check

We can verify this is correct by converting back: $\log_{16}(256) = 2$ means $16^2 = 256$, which matches our original equation.

Worked Example 2: Converting from logarithmic to exponential form

Question: Rewrite $\log_3(81) = 4$ in exponential form.

Strategy

We use the equivalence: if $\log_a(x) = y$, then $a^y = x$.

Working

Start with the logarithmic form:

$\log_a(x) = y$

Write the specific equation:

$\log_3(81) = 4$

This tells us that base $a = 3$, argument $x = 81$, and exponent $y = 4$.

Now convert to exponential form:

$a^y = x$

Substitute our values:

$3^4 = 81$

Check

To verify: $3^4 = 3 \times 3 \times 3 \times 3 = 81$. This confirms our result.

Worked Example 3: Evaluating logarithms without technology

Question: Evaluate these logarithms without technology:

Part a: $\log_4(64)$

Strategy

Set the logarithm equal to $y$ and convert to exponential form. Then express both sides with the same base.

Working

Let $\log_4(64) = y$

Convert to exponential form:

$4^y = 64$

Express 64 as a power of 4:

$4^y = 4^3$

Since the bases are equal, the exponents must be equal:

$y = 3$

Therefore, $\log_4(64) = 3$.

Part b: $\log_9\left(\frac{1}{81}\right)$

Strategy

Set the logarithm equal to $y$ and convert to exponential form. Use negative exponents for fractions.

Working

Let $\log_9\left(\frac{1}{81}\right) = y$

Convert to exponential form:

$9^y = \frac{1}{81}$

Express $\frac{1}{81}$ as a power of 9 (recall that $\frac{1}{a^n} = a^{-n}$):

$9^y = 9^{-2}$

Since the bases are equal, the exponents must be equal:

$y = -2$

Therefore, $\log_9\left(\frac{1}{81}\right) = -2$.

Worked Example 4: Estimating and using technology

Question: Consider $\log(7500)$:

Part a: Determine between which two consecutive integers $\log(7500)$ lies.

Strategy

Test powers of 10 to find bounds for 7500.

Working

$10^3 = 1000$ (test power of 3)

$10^4 = 10000$ (test power of 4)

Since $1000 < 7500 < 10000$, we have:

$\log(1000) < \log(7500) < \log(10000)$

This means:

$3 < \log(7500) < 4$

Therefore, $\log(7500)$ lies between 3 and 4.

Part b: Evaluate $\log(7500)$ using technology, rounded to four decimal places.

Working

Using a calculator:

$\log(7500) \approx 3.8751$

This confirms our estimate that the value lies between 3 and 4, and is closer to 4 than to 3.

Worked Example 5: Converting natural logarithms

Question: Convert these natural logarithms to exponential form:

Part a: $\ln(10) = y$

Strategy

Use the definition: if $\ln(x) = y$, then $e^y = x$.

Working

Write the natural logarithm form:

$\ln(x) = y$

Write the equation to compare:

$\ln(10) = y$

This shows that argument $x = 10$.

Convert to exponential form:

$e^y = x$

Substitute $x = 10$:

$e^y = 10$

This means: "e raised to the power y equals 10".

Part b: $\ln(e^3) = 3$

Strategy

Apply the inverse property: $\ln(e^y) = y$.

Working

Write the natural logarithm form:

$\ln(x) = y$

Write the equation to compare:

$\ln(e^3) = 3$

This shows that $x = e^3$ and $y = 3$.

Convert to exponential form:

$e^y = x$

Substitute values:

$e^3 = e^3$

This verifies the equation and demonstrates the inverse relationship between $e^x$ and $\ln(x)$.