Logarithms Across Subjects - Class 9 Advanced Maths

Recall the definition: $b^x = y \iff \log_b(y) = x$.

(a) Solution: $\mathbf{\log_5(625) = 4}$

(b) Solution: $\mathbf{\log_{10}(0.01) = -2}$

(c) Solution: $\mathbf{\log_7(1) = 0}$

(d) Solution: $\mathbf{\log_8(8) = 1}$

Method 1 (Using Quotient Rule):
$\log_2(16) - \log_2(4) = \log_2\left(\frac{16}{4}\right) = \log_2(4) = \log_2(2^2) = 2\log_2(2) = \mathbf{2}$.

Method 2 (Evaluating directly):
- Since $2^4 = 16$, $\log_2(16) = 4$.
- Since $2^2 = 4$, $\log_2(4) = 2$.
- Therefore, $\log_2(16) - \log_2(4) = 4 - 2 = \mathbf{2}$.

(a) Solution: Let $\log_2(256) = x \Rightarrow 2^x = 256 = 2^8 \Rightarrow \mathbf{x = 8}$.

(b) Solution: Let $\log_4(16) = x \Rightarrow 4^x = 16 = 4^2 \Rightarrow \mathbf{x = 2}$.

(c) Solution: Let $\log_5(125) = x \Rightarrow 5^x = 125 = 5^3 \Rightarrow \mathbf{x = 3}$.

(d) Solution: Let $\log_{10}(0.001) = x \Rightarrow 10^x = 0.001 = 10^{-3} \Rightarrow \mathbf{x = -3}$.

(a) Solution: $\log_2(21) = \log_2(7 \times 3) = \log_2(7) + \log_2(3) = \mathbf{p + q}$.

(b) Solution: $\log_2(49) = \log_2(7^2) = 2\log_2(7) = \mathbf{2p}$.

(c) Solution: $\log_2\left(\frac{7}{3}\right) = \log_2(7) - \log_2(3) = \mathbf{p - q}$.

(d) Solution: $\log_2(63) = \log_2(9 \times 7) = \log_2(3^2) + \log_2(7) = 2\log_2(3) + \log_2(7) = \mathbf{2q + p}$.

(a) Solution:
The brightness ratio is $100$.
$\text{Magnitude Difference} = 2.5\log_{10}(100) = 2.5(2) = \mathbf{5\text{ magnitudes}}.$

(b) Solution:
From $\text{pH} = -\log_{10}[\text{H}^+]$, we get $[\text{H}^+] = 10^{-\text{pH}}$.
- Acidity of first solution: $[\text{H}^+]_1 = 10^{-3}\text{ M}$.
- Acidity of second solution: $[\text{H}^+]_2 = 10^{-6}\text{ M}$.
- Ratio: $\frac{[\text{H}^+]_1}{[\text{H}^+]_2} = \frac{10^{-3}}{10^{-6}} = 10^3 = \mathbf{1000\text{ times}}$.
So, the first solution is **1000 times** more acidic.

(c) Solution:
Difference in Richter magnitudes: $9 - 4 = 5$ units.
Since Richter scale is base-10: difference of 5 units represents: $10^5 = \mathbf{100,000\text{ times stronger}}.$

(a) Answer: True
The logarithmic function is one-to-one. Therefore, equal outputs imply equal inputs.

(b) Answer: True
By the Change of Base formula: $\log_8(e) = \frac{\log_e(e)}{\log_e(8)} = \frac{1}{\ln(8)}$ (since $\ln(8) = \log_e(8)$ and $\log_e(e) = 1$).

(c) Answer: False
In real-number algebra, $b^y = x$. If base $b > 0$, then $b^y$ must always be positive. Thus, $x$ cannot be negative.

(d) Answer: False
The quotient rule is: $\log_b(M/N) = \log_b(M) - \log_b(N)$. There is no expansion rule for subtraction inside the logarithm.

(e) Answer: False
The base $b$ of a logarithm is strictly constrained to be positive and not equal to 1 ($b > 0, b \neq 1$).

Let's approximate each logarithmic term individually without using a calculator:

Step 1: Approximate $\log_4(9)$
- We know $4^1 = 4$ and $4^2 = 16$. Since $4 < 9 < 16$, we have $1 < \log_4(9) < 2$.
- Specifically, since $\sqrt{4^3} = \sqrt{64} = 8$, we know $4^{1.5} = 8$.
- Since $9 > 8$, we must have $\log_4(9) > 1.5$. Indeed, $\log_4(9) \approx 1.58$.

Step 2: Approximate $\log_9(28)$
- We know $9^1 = 9$ and $9^2 = 81$. Since $9 < 28 < 81$, we have $1 < \log_9(28) < 2$.
- Specifically, since $\sqrt{9^3} = \sqrt{729} = 27$, we know $9^{1.5} = 27$.
- Since $28 > 27$, we must have $\log_9(28) > 1.5$. Indeed, $\log_9(28) \approx 1.52$.

Step 3: Combine and find greatest integer
- The sum $\log_4(9) + \log_9(28)$ is slightly greater than $1.5 + 1.5 = 3$.
- Let's verify if the sum is less than 4:
- $\log_4(9) \approx 1.58$ and $\log_9(28) \approx 1.52$.
- Sum $\approx 1.58 + 1.52 = 3.10$.
- The number lies between $3$ and $4$.
- The greatest integer that is strictly less than this number is **3**.

Step 1: Solve for $x$ and $y$
From the definition of logarithms:
- $2^x = 5 \Rightarrow x = \log_2(5)$.
- We are also given: $y = \log_2(5)$.
Therefore, $x = y = \log_2(5)$.

Step 2: Approximate the value of $\log_2(5)$
- We know $2^2 = 4$ and $2^3 = 8$.
- Let's use the approximation $\log_2(5) \approx 2.32$.

Step 3: Evaluate $(x + 5y)$
Since $x = y$:
$x + 5y = y + 5y = 6y = 6 \log_2(5).$
Substituting the approximate value:
$6 \times 2.3219 \approx \mathbf{13.93}.$