Question: b) Evaluate and simplify the following expressions.
i. Simplify completely log₂(128) + 2log₂(4) − log₂(8) (02 Marks)
ii. Solve for x,
log₅(x + 9) − log₅(x − 1) = 2. State any restrictions on x. (02 Marks)
iii. A chemical reaction follows the equation 3^(2x+1) = 27^(x−1) × 81. Using laws of indices
and logarithms, determine the value of x and verify your result. (03 Marks)
1. **Problem Statement:**
Evaluate and simplify the following expressions:
i. Simplify completely $$\log_2(128) + 2\log_2(4) - \log_2(8)$$
ii. Solve for $$x$$ in $$\log_5(x + 9) - \log_5(x - 1) = 2$$ and state restrictions on $$x$$.
iii. Solve for $$x$$ in $$3^{2x+1} = 27^{x-1} \times 81$$ using laws of indices and logarithms, then verify.
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2. **Important formulas and rules:**
- $$\log_b(a) + \log_b(c) = \log_b(ac)$$
- $$k \log_b(a) = \log_b(a^k)$$
- $$\log_b(a) - \log_b(c) = \log_b\left(\frac{a}{c}\right)$$
- Laws of indices: $$a^m \times a^n = a^{m+n}$$
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### i. Simplify $$\log_2(128) + 2\log_2(4) - \log_2(8)$$
1. Express each term as powers of 2:
$$128 = 2^7, \quad 4 = 2^2, \quad 8 = 2^3$$
2. Rewrite logarithms:
$$\log_2(128) = 7$$
$$2\log_2(4) = 2 \times 2 = 4$$
$$\log_2(8) = 3$$
3. Substitute back:
$$7 + 4 - 3 = 8$$
**Answer:** $$8$$
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### ii. Solve $$\log_5(x + 9) - \log_5(x - 1) = 2$$
1. Use log subtraction rule:
$$\log_5\left(\frac{x+9}{x-1}\right) = 2$$
2. Convert to exponential form:
$$\frac{x+9}{x-1} = 5^2 = 25$$
3. Solve for $$x$$:
$$x + 9 = 25(x - 1)$$
$$x + 9 = 25x - 25$$
$$9 + 25 = 25x - x$$
$$34 = 24x$$
$$x = \frac{34}{24} = \frac{17}{12}$$
4. **Restrictions:**
- $$x + 9 > 0 \Rightarrow x > -9$$
- $$x - 1 > 0 \Rightarrow x > 1$$
Since $$x = \frac{17}{12} \approx 1.416 > 1$$, it satisfies restrictions.
**Answer:** $$x = \frac{17}{12}$$
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### iii. Solve $$3^{2x+1} = 27^{x-1} \times 81$$
1. Express all terms as powers of 3:
$$27 = 3^3, \quad 81 = 3^4$$
2. Rewrite equation:
$$3^{2x+1} = (3^3)^{x-1} \times 3^4$$
3. Simplify right side using power of a power:
$$3^{2x+1} = 3^{3(x-1)} \times 3^4 = 3^{3x - 3 + 4} = 3^{3x + 1}$$
4. Since bases are equal, equate exponents:
$$2x + 1 = 3x + 1$$
5. Solve for $$x$$:
$$2x + 1 = 3x + 1$$
$$2x + 1 - 1 = 3x + 1 - 1$$
$$2x = 3x$$
$$2x - 3x = 0$$
$$-x = 0$$
$$x = 0$$
6. **Verification:**
Substitute $$x=0$$ into original equation:
Left side:
$$3^{2(0)+1} = 3^1 = 3$$
Right side:
$$27^{0-1} \times 81 = 27^{-1} \times 81 = \frac{1}{27} \times 81 = 3$$
Both sides equal 3, so $$x=0$$ is correct.
**Answer:** $$x=0$$