Exponents and Logarithm : Lesson

Objective

• Understand and apply the laws of exponents
• Learn logarithmic laws and their connection to exponents
• Use natural logarithms and base-changing rules

Laws of Exponents

• $a^m \cdot a^n = a^{m+n}$

• $a^0 = 1,\ a \ne 0$

• $\frac{a^m}{a^n} = a^{m-n}$

• $a^{-n} = \frac{1}{a^n}$ and $\frac{1}{a^{-n}} = a^n$

• $(a^m)^n = a^{mn}$

• $a^{1/n} = \sqrt[n]{a}$

• $(ab)^n = a^n b^n$

• $a^{m/n} = \sqrt[n]{a^m} = (\sqrt[n]{a})^m$

• $\left(\frac{a}{b}\right)^n = \frac{a^n}{b^n}$

• If $a^x = a^k$, then $x = k$ (when bases are the same, we equate indices).

Definition of a Logarithm

If $a^x = b$, where $a \ne 1$ and $a > 0,\ b > 0$, then $x = \log_a b$

Also: $x = \log_a (a^x)$ and $x = a^{\log_a x}$ (provided $x > 0$)

The natural logarithm uses base $e$ and is written as: $\ln x \equiv \log_e x$

Laws of Logarithms

• $\log_a(xy) = \log_a x + \log_a y$

• $\log_a\left(\frac{x}{y}\right) = \log_a x – \log_a y$

• $\log_a(x^m) = m \log_a x$

• $\log_a 1 = 0$

• $\log_a a = 1$

Natural Logarithm Laws (base $e$)

• $\ln(xy) = \ln x + \ln y$

• $\ln\left(\frac{x}{y}\right) = \ln x – \ln y$

• $\ln(x^m) = m \ln x$

• $\ln 1 = 0$

• $\ln e = 1$

Change of Base Formula

$\log_a x = \frac{\log_b x}{\log_b a}$ (Valid for any $b > 0,\ b \ne 1$)

Example 1 : Simplify $2\log_a 6 + \log_a 2 – \log_a 12$

Simplify: $2\log_a 6 + \log_a 2 – \log_a 12$

(Apply power rule) $\log_a 6^2 + \log_a 2 – \log_a 12$

$= \log_a 36 + \log_a 2 – \log_a 12$

(Apply product rule) $= \log_a (36 \cdot 2) – \log_a 12$

$= \log_a 72 – \log_a 12$

(Apply quotient rule) $= \log_a \left(\frac{72}{12}\right) = \log_a 6$

Final Answer ${\log_a 6}$

Example 2: Simplify: $\log_3 x + 2\log_3 y – 3\log_3 t$

Simplify: $\log_3 x + 2\log_3 y – 3\log_3 t$

(Apply power rule) $= \log_3 x + \log_3 y^2 – \log_3 t^3$

(Apply product and quotient rules) $= \log_3 (x \cdot y^2) – \log_3 t^3$

$= \log_3 \left(\frac{x y^2}{t^3}\right)$

Final Answer : ${\log_3 \left(\frac{x y^2}{t^3}\right)}$

Example 3: Simplify: $3\log_a(x+2) – \log_a(3x^2 – 12) + \log_a(x – 2)$

Simplify: $3\log_a(x+2) – \log_a(3x^2 – 12) + \log_a(x – 2)$

(Factor the middle term) $= 3\log_a(x+2) – \log_a[3(x+2)(x-2)] + \log_a(x – 2)$

(Apply power and log rules) $= \log_a[(x+2)^3] – \log_a[3(x+2)(x-2)] + \log_a(x – 2)$

(Group and apply product/quotient rules) $= \log_a \left(\frac{(x+2)^3 (x – 2)}{3(x+2)(x – 2)}\right)$

(Simplify) $= \log_a \left(\frac{(x+2)^2}{3}\right)$

Final Answer: ${\log_a \left(\frac{(x+2)^2}{3}\right)}$

Example 4: Given: $\log_2 y = 3\log_2 q + 5\log_2 x – \frac{1}{2} \log_2 p$. Expressing $y$ in Terms of $x$ Using Logarithmic Rules

Given: $\log_2 y = 3\log_2 q + 5\log_2 x – \frac{1}{2} \log_2 p$

Use power and product rules: $= \log_2(q^3) + \log_2(x^5) – \log_2(\sqrt{p})$

Combine using product and quotient rules: $= \log_2 \left( \frac{q^3 x^5}{\sqrt{p}} \right)$

Since $\log_2 y = \log_2 \left( \frac{q^3 x^5}{\sqrt{p}} \right)$,

we equate arguments: $y = \frac{q^3 x^5}{\sqrt{p}}$

Final Answer : ${y = \frac{q^3 x^5}{\sqrt{p}}}$

Example 5: Solve the following logarithmic equations without using a calculator. Do not use the final line boxing method.

[a] $\log_a x + 2\log_a 5 = \log_a 225$

[b] $\log_5(x+1) + \log_5(x-3) = 1$

[c] $\log_{16}(x+2) – \log_{16}(x-6) = \frac{1}{2}$

[d] $\log_7(2x+5) – \log_7(x-5) = \log_7\left(\frac{x}{2}\right)$

[e] $\log_{10}(x+2) – \log_{10}x = 2\log_{10}4$

[a] $\log_a x + 2\log_a 5 = \log_a 225$

Use the power rule: $\log_a x + \log_a 25 = \log_a 225$

Now use the product rule: $\log_a(25x) = \log_a 225$

Equating arguments: $25x = 225$

Solve: $x = \frac{225}{25} = 9$

[b] $\log_5(x+1) + \log_5(x-3) = 1$

Use the product rule: $\log_5((x+1)(x-3)) = 1$

Simplify: $\log_5(x^2 – 2x – 3) = 1$

Rewrite in exponential form: $x^2 – 2x – 3 = 5^1 = 5$

Solve: $x^2 – 2x – 8 = 0$

Factor: $(x – 4)(x + 2) = 0 \Rightarrow x = 4$ or $x = -2$

Check domain: $x = -2$ invalid (log of negative),

Valid solution: $x = 4$

[c] $\log_{16}(x + 2) – \log_{16}(x – 6) = \frac{1}{2}$

Use the quotient rule: $\log_{16} \left( \frac{x + 2}{x – 6} \right) = \frac{1}{2}$

Rewrite in exponential form: $\frac{x + 2}{x – 6} = 16^{1/2} = 4$

Now solve: $\frac{x + 2}{x – 6} = 4$

Cross-multiply: $x + 2 = 4(x – 6) = 4x – 24$

$3x = 26 \Rightarrow x = \frac{26}{3}$

[d] $\log_7(2x + 5) – \log_7(x – 5) = \log_7\left( \frac{x}{2} \right)$

Use the quotient rule on the left: $\log_7\left( \frac{2x + 5}{x – 5} \right) = \log_7\left( \frac{x}{2} \right)$

Equating arguments: $\frac{2x + 5}{x – 5} = \frac{x}{2}$

Cross-multiply: $2(2x + 5) = x(x – 5)$

$4x + 10 = x^2 – 5x$

Bring all to one side: $x^2 – 9x – 10 = 0$

Factor: $(x – 10)(x + 1) = 0$

Check domain: $x = -1$ invalid, $x = 10$ valid

Valid solution: $x = 10$

[e] $\log_{10}(x + 2) – \log_{10} x = 2\log_{10} 4$

Apply quotient rule on LHS and power rule on RHS:

$\log_{10}\left( \frac{x + 2}{x} \right) = \log_{10}(4^2) = \log_{10}(16)$

Equating arguments: $\frac{x + 2}{x} = 16$

Simplify: $\frac{x + 2}{x} = 16 \Rightarrow x + 2 = 16x \Rightarrow 15x = 2 \Rightarrow x = \frac{2}{15}$

Example 6: Solve the equation: $7 \cdot 3^{x+1} = 2 + \frac{3}{3^x}.$ Give your answer in the form $x = a – \log_3 b$, where $a, b \in \mathbb{Z}$.

Solution:

Step 1: Rewrite all terms with powers of 3

We can rewrite $3^{x+1} = 3 \cdot 3^x$, and the denominator $\frac{3}{3^x} = 3^{1 – x}$.

So the equation becomes: $7 \cdot 3 \cdot 3^x = 2 + 3^{1 – x}$

That simplifies to: $21 \cdot 3^x = 2 + 3^{1 – x}$

Step 2: Let $y = 3^x$

Then the equation becomes: $21y = 2 + \frac{3}{y}$

Multiply both sides by $y$: $21y^2 = 2y + 3$

Bring all terms to one side: $21y^2 – 2y – 3 = 0$

Step 3: Solve the quadratic

Use the quadratic formula: $y = \frac{-(-2) \pm \sqrt{(-2)^2 – 4(21)(-3)}}{2 \cdot 21}$

$y = \frac{2 \pm \sqrt{4 + 252}}{42} = \frac{2 \pm \sqrt{256}}{42} = \frac{2 \pm 16}{42}$

So: $y = \frac{18}{42} = \frac{3}{7}$ or $y = \frac{-14}{42} = -\frac{1}{3}$

We reject the negative value since $y = 3^x > 0$

Step 4: Back-substitute

We had $y = 3^x = \frac{3}{7}$

Take logs on both sides:

$x = \log_3\left(\frac{3}{7}\right) = \log_3 3 – \log_3 7 = 1 – \log_3 7$

Final Answer: $x = 1 – \log_3 7$

Example 7: Find the exact values of $x$ satisfying the equation. $(3^x)(4^{2x+1}) = 6^{x+2}$, giving your answer in the form $x = \frac{\ln a}{\ln b}$, where $a,b \in \mathbb{Z}$.

Rewrite all terms using prime bases

$4 = 2^2,\quad 6 = 2 \cdot 3$

So the equation becomes: $3^x \cdot (2^2)^{2x+1} = (2 \cdot 3)^{x+2}$

$3^x \cdot 2^{4x+2} = 2^{x+2} \cdot 3^{x+2}$

Group like bases and simplify

Move all terms to one side:

$\frac{3^x \cdot 2^{4x+2}}{2^{x+2} \cdot 3^x} = \frac{2^{x+2} \cdot 3^{x+2}}{2^{x+2} \cdot 3^x}$

This simplifies to: $2^{(4x+2)-(x+2)} = 3^{(x+2)-x}$

$2^{3x} = 3^2$

Apply natural logarithm to both sides

$\ln(2^{3x}) = \ln(3^2)$

$3x \ln 2 = 2 \ln 3$

Solve for $x$

$x = \frac{2 \ln 3}{3 \ln 2}$

$\quad = \frac{\ln 9}{\ln 8}$

Final Answer ${x = \frac{\ln 9}{\ln 8}}$

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