Scottish Highers Maths Exponential and Logs: 14. (a) Evaluate $ ext{log}

14. (a) Evaluate $ ext{log}_{8} 4 + 2 ext{log}_{8} 5$ - Scottish Highers Maths - Question 14 - 2019

Question 14

$14.-(a)-Evaluate-$-ext{log}_{8}-4-+-2--ext{log}_{8}-5$-Scottish Highers Maths-Question 14-2019.png$

14. (a) Evaluate $ ext{log}_{8} 4 + 2 ext{log}_{8} 5$. (b) Solve $ ext{log}_{3}(7x-2) - ext{log}_{3} 3 = 5$, $x geq 1$.

Worked Solution & Example Answer:14. (a) Evaluate $ ext{log}_{8} 4 + 2 ext{log}_{8} 5$ - Scottish Highers Maths - Question 14 - 2019

Step 1

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Answer

To evaluate this expression, we will first apply the logarithmic identities.

Using the property of logarithms that states: $ext{log}_{b} a^n = n \cdot \text{log}_{b} a$ we simplify $2 \text{log}_{8} 5$ : $2 \text{log}_{8} 5 = \text{log}_{8} 5^2 = \text{log}_{8} 25$
Now substitute back into the expression:
$\text{log}_{8} 4 + \text{log}_{8} 25$
Using the property that states: $\text{log}_{b} a + \text{log}_{b} c = \text{log}_{b} (a \times c)$ we can combine the two: $\text{log}_{8} (4 \times 25) = \text{log}_{8} 100$
Next, we can express $100$ as a power of $8$ . We know that: $8^{2} = 64 \quad \text{and} \quad 8^{2.5} \approx 100$

Thus, we conclude: $\text{log}_{8} 100 \approx 2.5$ .

Step 2

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Answer

First, we can use the property of logarithms: $\text{log}_{b} a - \text{log}_{b} c = \text{log}_{b} \left( \frac{a}{c} \right)$ to simplify the left-hand side: $\text{log}_{3} \left( \frac{7x - 2}{3} \right) = 5$
Next, we convert the logarithmic equation to exponential form:
$\frac{7x - 2}{3} = 3^5$ $\frac{7x - 2}{3} = 243$
Multiply both sides by 3:
$7x - 2 = 729$
Add 2 to both sides:
$7x = 731$
Finally, divide by 7 to solve for $x$ :
$x = \frac{731}{7} \approx 104.43$

Since $x \geq 1$ , this solution is valid.