Scottish Highers Maths Exponential and Logs: 7. (a) Evaluate $\log

7. (a) Evaluate $\log_2 5 + \log_2 \frac{1}{40}$ - Scottish Highers Maths - Question 7 - 2023

Question 7

$7.-(a)-Evaluate-$\log_2-5-+-\log_2-\frac{1}{40}$-Scottish Highers Maths-Question 7-2023.png$

7. (a) Evaluate $\log_2 5 + \log_2 \frac{1}{40}$. (b) Given that $a \in \mathbb{R}$ and that $\log_a u$ is negative, state the range of possible values of $a$.

Worked Solution & Example Answer:7. (a) Evaluate $\log_2 5 + \log_2 \frac{1}{40}$ - Scottish Highers Maths - Question 7 - 2023

Step 1

Evaluate $\log_2 5 + \log_2 \frac{1}{40}$

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Answer

To evaluate the expression, we can use the properties of logarithms. The property states that the sum of logarithms can be combined into a single logarithm:

$\log_b x + \log_b y = \log_b (x \cdot y)$

Applying this, we get:

$\log_2 5 + \log_2 \frac{1}{40} = \log_2 \left( 5 \cdot \frac{1}{40} \right) = \log_2 \left( \frac{5}{40} \right) = \log_2 \left( \frac{1}{8} \right)$

Next, we recognize that ( \frac{1}{8} = 2^{-3} ), which means:

$\log_2 \left( \frac{1}{8} \right) = \log_2 (2^{-3}) = -3$

Thus, the final answer is:

$-3$

Step 2

Given that $a \in \mathbb{R}$ and that $\log_a u$ is negative, state the range of possible values of $a$.

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Answer

For the logarithm $\log_a u$ to be negative, the base $a$ must be determined. The logarithm is negative when:

$0 < u < 1$ with $a > 1$ .
$u > 1$ with $0 < a < 1$ .

Therefore, the possible ranges for $a$ are:

$0 < a < 1 \text{ or } a > 1$

7. (a) Evaluate $\log_2 5 + \log_2 \frac{1}{40}$ - Scottish Highers Maths - Question 7 - 2023

Worked Solution & Example Answer:7. (a) Evaluate $\log_2 5 + \log_2 \frac{1}{40}$ - Scottish Highers Maths - Question 7 - 2023

Evaluate $\log_2 5 + \log_2 \frac{1}{40}$

Given that $a \in \mathbb{R}$ and that $\log_a u$ is negative, state the range of possible values of $a$.

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