Let's solve each part of the question step by step.
### Part (a)
#### (i) [tex]\(\log_x 343 = 3\)[/tex]
The logarithmic equation can be written in exponential form:
[tex]\[
343 = x^3
\][/tex]
To solve for [tex]\(x\)[/tex], we take the cube root of both sides:
[tex]\[
x = \sqrt[3]{343}
\][/tex]
Since [tex]\(343 = 7^3\)[/tex], we have:
[tex]\[
x = 7
\][/tex]
#### (ii) [tex]\(\log_9 (4y - 3) = 2\)[/tex]
The logarithmic equation can be written in exponential form:
[tex]\[
9^2 = 4y - 3
\][/tex]
Simplifying:
[tex]\[
81 = 4y - 3
\][/tex]
Adding 3 to both sides:
[tex]\[
84 = 4y
\][/tex]
Dividing both sides by 4:
[tex]\[
y = \frac{84}{4} = 21
\][/tex]
### Part (b)
To solve [tex]\(\log_q 5 + 6 \log_5 q = 5\)[/tex]:
Let [tex]\(\log_q 5 = a\)[/tex]. This implies that [tex]\(\log_5 q = \frac{1}{a}\)[/tex].
Substitute these into the equation:
[tex]\[
a + 6 \cdot \frac{1}{a} = 5
\][/tex]
Multiplying through by [tex]\(a\)[/tex] to clear the fraction:
[tex]\[
a^2 + 6 = 5a
\][/tex]
Rearranging gives:
[tex]\[
a^2 - 5a + 6 = 0
\][/tex]
This is a quadratic equation which can be solved using the quadratic formula [tex]\(a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex]:
Here, [tex]\(a = 1\)[/tex], [tex]\(b = -5\)[/tex], and [tex]\(c = 6\)[/tex]. Solving for [tex]\(a\)[/tex]:
[tex]\[
a = \frac{5 \pm \sqrt{25 - 24}}{2} = \frac{5 \pm 1}{2}
\][/tex]
This gives two possible solutions:
[tex]\[
a = 3 \quad \text{or} \quad a = 2
\][/tex]
Since [tex]\(\log_q 5 = a\)[/tex], we need the correct value of [tex]\(q\)[/tex]:
For [tex]\(a = 3\)[/tex]:
[tex]\[
\log_q 5 = 3 \implies q = 5^{1/3}
\][/tex]
For [tex]\(a = 2\)[/tex]:
[tex]\[
\log_q 5 = 2 \implies q = 5^{1/2}
\][/tex]
Comparing numerical values to 3 significant figures, the correct solution is:
[tex]\[
q \approx 1.70997594667670
\][/tex]
### Part (c)
We need to show that:
[tex]\[
x \log_2(x^5) - \log_2(x^2) \equiv (5x - 2) \log_2(x)
\][/tex]
Using logarithm properties:
[tex]\[
\log_2(x^5) = 5 \log_2(x)
\][/tex]
[tex]\[
\log_2(x^2) = 2 \log_2(x)
\][/tex]
Substitute these into the left-hand side expression:
[tex]\[
x \cdot 5 \log_2(x) - 2 \log_2(x) = 5x \log_2(x) - 2 \log_2(x)
\][/tex]
[tex]\[
= (5x - 2) \log_2(x)
\][/tex]
Hence, the identity holds true.
### Part (d)
Using the result from part (c), we solve the equation:
[tex]\[
x \log_2(x^5) - \log_2(x^2) = 20x - 8
\][/tex]
From part (c), we know:
[tex]\[
(5x - 2) \log_2(x) = 20x - 8
\][/tex]
Divide both sides by [tex]\(\log_2(x)\)[/tex]:
[tex]\[
5x - 2 = \frac{20x - 8}{\log_2(x)}
\][/tex]
Solving this equation numerically yields the values:
[tex]\[
x = \frac{2}{5}, \quad x = 16
\][/tex]
Thus, the solutions are [tex]\(x = \frac{2}{5}\)[/tex] and [tex]\(x = 16\)[/tex].
