Exercise 2.2 — Ideal Textbook Solutions

(a) Solution:
Apply the Power Rule first: $2\log(7) = \log(7^2) = \log(49)$.
Now apply the Product Rule: $\log(2) + \log(49) = \log(2 \times 49) = \mathbf{\log(98)}$.

(b) Solution:
Apply the Quotient Rule sequentially:
$\log_3(8) - \log_3(5) - \log_3(4) = \log_3\left(\frac{8}{5}\right) - \log_3(4) = \log_3\left(\frac{8}{5 \times 4}\right) = \log_3\left(\frac{8}{20}\right) = \mathbf{\log_3\left(\frac{2}{5}\right)}$.

(c) Solution:
Simplify exponent: $2\log(3) = \log(3^2) = \log(9)$.
Combine: $\log(5) + \log(9) - \log(15) = \log(45) - \log(15) = \log\left(\frac{45}{15}\right) = \mathbf{\log(3)}$.

(d) Solution:
Express 5 as a logarithm with base 2: $5 = \log_2(2^5) = \log_2(32)$.
Combine: $\log_2(3) - \log_2(32) = \mathbf{\log_2\left(\frac{3}{32}\right)}$.

(e) Solution:
Apply power rule: $\frac{1}{2}\log_3(9) = \log_3(9^{1/2}) = \log_3(3) = \mathbf{1}$.

(f) Solution:
Apply power rule: $\log_4(3^{1/2}) - \log_4(4^3) = \log_4(\sqrt{3}) - \log_4(64) = \mathbf{\log_4\left(\frac{\sqrt{3}}{64}\right)}$.

(a) Solution: Since $11^2 = 121$, we have: $\log_{11}(121) = 2$.

(b) Solution: Since $7^0 = 1$, we have: $\log_7(1) = 0$.

(c) Solution: Since $5^4 = 625$, we have: $\log_5(625) = 4$.

(d) Solution: Since $8^1 = 8$, we have: $\log_8(8) = 1$.

(e) Solution: Standard base is 10. Since $10^3 = 1000$, we have: $\log(1000) = 3$.

(a) Solution:
$\log_2(15) = \log_2(3 \times 5) = \log_2(3) + \log_2(5) = \mathbf{p + q}$.

(b) Solution:
$\log_2(45) = \log_2(3^2 \times 5) = 2\log_2(3) + \log_2(5) = \mathbf{2p + q}$.

(c) Solution:
$\log_2\left(\frac{5}{3}\right) = \log_2(5) - \log_2(3) = \mathbf{q - p}$.

(d) Solution:
$\log_2(10) = \log_2(2 \times 5) = \log_2(2) + \log_2(5) = 1 + \log_2(5) = \mathbf{1 + q}$ (since $\log_2(2) = 1$).

(a) Answer: True (Conditional implication)
By the one-to-one property of exponential functions, if the bases are equal, the exponents must be equal. Thus, the implication itself is logically true, even though the resulting equation $x+1=x+2 \Rightarrow 1=2$ has no real solution.

(b) Answer: False
$\log(x) + \log(1) = \log(x \times 1) = \log(x)$. Since $\log(x+1) \neq \log(x)$, this statement is False.

(c) Answer: True
By the Power Rule: $\log_b(b^3) = 3\log_b(b) = 3(1) = 3$.

(d) Answer: True
The base $b$ of a logarithm is strictly defined as $b > 0$ and $b \neq 1$. If the base were 1, then $1^y = x$, which has no unique solution for $x \neq 1$. Thus, base 1 is not defined.

Let's simplify the system of equations by letting $\log_{2026}(x) = X$,   $\log_{2026}(y) = Y$,   and $\log_{2026}(z) = Z$.

We are given:
(1) $X + Y = a$
(2) $Y + Z = b$
(3) $Z + X = c$

Step 1: Sum all three equations:
$2(X + Y + Z) = a + b + c$
$\Rightarrow X + Y + Z = \frac{a+b+c}{2}$  --- (4)

Step 2: Solve for $X, Y, Z$ individually:
- Subtract (2) from (4): $X = (X+Y+Z) - (Y+Z) = \frac{a+b+c}{2} - b = \frac{a+c-b}{2}$.
- Subtract (3) from (4): $Y = (X+Y+Z) - (Z+X) = \frac{a+b+c}{2} - c = \frac{a+b-c}{2}$.
- Subtract (1) from (4): $Z = (X+Y+Z) - (X+Y) = \frac{a+b+c}{2} - a = \frac{b+c-a}{2}$.

Step 3: Convert back to exponential form to find $x, y, z$:
- $\log_{2026}(x) = \frac{a+c-b}{2} \Rightarrow \mathbf{x = 2026^{\frac{a+c-b}{2}}}$

- $\log_{2026}(y) = \frac{a+b-c}{2} \Rightarrow \mathbf{y = 2026^{\frac{a+b-c}{2}}}$

- $\log_{2026}(z) = \frac{b+c-a}{2} \Rightarrow \mathbf{z = 2026^{\frac{b+c-a}{2}}}$