Exercise 2.4 — Ideal Textbook Solutions

(a) Solution:
Convert to exponential form:
$$2x - 5 = 3^2 = 9 \Rightarrow 2x = 14 \Rightarrow \mathbf{x = 7}.$$
*Check:* $2(7)-5 = 9 > 0$ (defined). Valid!

(b) Solution:
Apply product rule:
$$\log_7(3x × 2) = \log_7(24) \Rightarrow \log_7(6x) = \log_7(24)$$
Equate arguments:
$$6x = 24 \Rightarrow \mathbf{x = 4}.$$
*Check:* $3(4) = 12 > 0$ (defined). Valid!

(c) Solution:
Apply product rule:
$$\log_5[(x-3)(x+1)] = 1 \Rightarrow (x-3)(x+1) = 5^1$$
$$x^2 - 2x - 3 = 5 \Rightarrow x^2 - 2x - 8 = 0$$
$$(x-4)(x+2) = 0 \Rightarrow x = 4 or x = -2$$
*Check:*
- If $x = 4$: both $x-3 = 1 > 0$ and $x+1 = 5 > 0$ are positive. Valid!
- If $x = -2$: $x-3 = -5 < 0$ (undefined). Extraneous root.
Thus, the only valid solution is **$x = 4$**.

(d) Solution:
Convert to exponential form:
$$x^2 - 7 = 2^3 = 8 \Rightarrow x^2 = 15 \Rightarrow \mathbf{x = ±√15}.$$
*Check:* For both $±√15$, $x^2 - 7 = 15 - 7 = 8 > 0$ (defined). Valid!

(a) Solution:
Apply product rule:
$$\log_2[(x-3)(x+1)] = 5 \Rightarrow (x-3)(x+1) = 2^5 = 32$$
$$x^2 - 2x - 3 = 32 \Rightarrow x^2 - 2x - 35 = 0$$
$$(x-7)(x+5) = 0 \Rightarrow x = 7 or x = -5$$
*Check:* $x = 7$ makes arguments positive. $x = -5$ makes arguments negative. Thus, **$x = 7$**.

(b) Solution:
Power rule first:
$$\log_4(x^2) = \log_4(5x - 4) \Rightarrow x^2 = 5x - 4$$
$$x^2 - 5x + 4 = 0 \Rightarrow (x-4)(x-1) = 0 \Rightarrow \mathbf{x = 4, 1}.$$
*Check:* Both arguments are positive for $x=4$ and $x=1$. Both are valid!

(c) Solution:
Product rule:
$$\log_5[(x-2)(x+2)] = 1 \Rightarrow x^2 - 4 = 5 \Rightarrow x^2 = 9 \Rightarrow x = ± 3.$$
*Check:* $x = 3$ yields positive arguments ($1 > 0$, $5 > 0$). $x = -3$ yields negative arguments. Thus, **$x = 3$**.

(d) Solution:
Product rule:
$$\log₁₀[(x-2)(x+1)] = 1 \Rightarrow (x-2)(x+1) = 10^1 = 10$$
$$x^2 - x - 2 = 10 \Rightarrow x^2 - x - 12 = 0$$
$$(x-4)(x+3) = 0 \Rightarrow x = 4 or x = -3$$
*Check:* $x = 4$ yields positive arguments. $x = -3$ yields negative arguments. Thus, **$x = 4$**.

(a) Solution:
Convert to exponential form:
$$x^2 = 3x + 10 \Rightarrow x^2 - 3x - 10 = 0 \Rightarrow (x-5)(x+2) = 0$$
$$\Rightarrow x = 5 or x = -2$$
*Check base constraint:* The base $x$ must be strictly positive and not equal to 1. Thus, $x = -2$ is rejected. The only valid solution is **$x = 5$**.

(b) Solution:
Let $y = \log_3 x$. Then:
$$y^2 - 4y + 3 = 0 \Rightarrow (y-3)(y-1) = 0 \Rightarrow y = 3 or y = 1$$
- If $\log_3 x = 3 \Rightarrow x = 3^3 = 27$.
- If $\log_3 x = 1 \Rightarrow x = 3^1 = 3$.
Both values are positive. Thus, **$x = 3, 27$**.

(c) Solution:
Rewrite using power rule: $\log_2(x^3) = 3\log_2 x$. Let $y = \log_2 x$.
$$y^2 - 3y = 10 \Rightarrow y^2 - 3y - 10 = 0 \Rightarrow (y-5)(y+2) = 0 \Rightarrow y = 5 or y = -2$$
- If $\log_2 x = 5 \Rightarrow x = 2^5 = 32$.
- If $\log_2 x = -2 \Rightarrow x = 2^-2 = 14$.
Both values are positive. Thus, **$x = 32, 14$**.

(d) Solution:
Express 1000 as $10^3$:
$$(10^3)^{\log₁₀ x} = 10^{3\log₁₀ x} = 10^{\log₁₀(x^3)} = x^3$$
$$\Rightarrow x^3 = 8 \Rightarrow \mathbf{x = 2}.$$

(a) Solution:
Equate arguments directly:
$$x^2 - 1 = 2x - 1 \Rightarrow x^2 - 2x = 0 \Rightarrow x(x - 2) = 0 \Rightarrow x = 2 or x = 0$$
*Check:* If $x = 0$, $x^2 - 1 = -1 < 0$ (undefined). Thus, $x = 0$ is rejected. The only valid solution is **$x = 2$**.

(b) Solution:
$\log(5 × 2) = \log(x) \Rightarrow \log(10) = \log(x) \Rightarrow \mathbf{x = 10}$.

(c) Solution:
$x^2 = 4 \Rightarrow x = ± 2$. Since base $x > 0$, we have **$x = 2$**.

(d) Solution:
Change base to 10:
$$\log x\log 2 + \log x\log 5 = \log x$$
$$\log x (1\log 2 + 1\log 5 - 1) = 0$$
This implies $\log x = 0 \Rightarrow \mathbf{x = 1}$ (the only real solution).

Step 1: Simplify the product chain of logarithms using the base-change identity $\log_a(b) · \log_b(c) = \log_a(c)$:

$$x = \log_2(3) · \log_3(4) · \log_4(5) \dots \log₉₉(100)$$ $$\Rightarrow x = \log_2(100)$$

Wait, if the problem assumes all terms are base 10 chain properties (i.e. $\log₁₀(2)/\log₁₀(1) \dots$):

If the equation is indeed the standard chain product:
$$x = \log_2(100) ≈ 6.64$$
Let's check if the problem specifies that the sum evaluates to a clean integer. If the bases are sequentially linked, the product simplifies to $\log_2(100)$.