Answer :
Sure, let's solve these equations step by step.
### Part (a)
Equation: [tex]\(\log_x 27 + 2 \log_3 x = 7\)[/tex]
1. Change of Base Formula: Recall that [tex]\(\log_a b = \frac{\log b}{\log a}\)[/tex].
We can rewrite [tex]\(\log_x 27\)[/tex] and [tex]\(\log_3 x\)[/tex] using the change of base formula:
[tex]\[ \log_x 27 = \frac{\log 27}{\log x} \][/tex]
[tex]\[ \log_3 x = \frac{\log x}{\log 3} \][/tex]
2. Substitute these expressions into the equation:
[tex]\[ \frac{\log 27}{\log x} + 2 \cdot \frac{\log x}{\log 3} = 7 \][/tex]
3. Simplify: Combine terms over a common denominator.
[tex]\[ \frac{\log 27 \cdot \log 3 + 2 (\log x)^2}{\log x \cdot \log 3} = 7 \][/tex]
4. Log Value: Recall that [tex]\(27 = 3^3\)[/tex], thus [tex]\(\log 27 = \log (3^3) = 3 \log 3\)[/tex]. Substitute this into the equation:
[tex]\[ \frac{3 \log 3 \cdot \log 3 + 2 (\log x)^2}{\log x \cdot \log 3} = 7 \][/tex]
[tex]\[ \frac{3 (\log 3)^2 + 2 (\log x)^2}{\log x \cdot \log 3} = 7 \][/tex]
5. Combine and Rearrange to a standard quadratic form in terms of [tex]\(\log x\)[/tex]:
[tex]\[ 3 (\log 3)^2 + 2 (\log x)^2 = 7 \log x \cdot \log 3 \][/tex]
6. Solve the Quadratic Equation: Let [tex]\(y = \log x\)[/tex]. The equation becomes:
[tex]\[ 2y^2 - 7 (\log 3)y + 3 (\log 3)^2 = 0 \][/tex]
7. Factorize or Use the Quadratic Formula to find [tex]\(y\)[/tex] (i.e., [tex]\(\log x\)[/tex]):
[tex]\[ y = \frac{7 (\log 3) \pm \sqrt{(7 (\log 3))^2 - 4 \cdot 2 \cdot 3 (\log 3)^2}}{2 \cdot 2} \][/tex]
[tex]\[ y = \frac{7 \log 3 \pm \sqrt{49 (\log 3)^2 - 24 (\log 3)^2}}{4} \][/tex]
[tex]\[ y = \frac{7 \log 3 \pm 5 \log 3}{4} \][/tex]
This gives us two solutions:
[tex]\[ y = 3 (\log 3) \quad \text{and} \quad y = \frac{\log 3}{2} \][/tex]
8. Find [tex]\(x\)[/tex]:
[tex]\[ \log x = 3 (\log 3) \Rightarrow x = 3^3 = 27 \][/tex]
[tex]\[ \log x = \frac{\log 3}{2} \Rightarrow x = 3^{1/2} = \sqrt{3} \][/tex]
Thus, the solutions are [tex]\( x = 27 \)[/tex] and [tex]\( x = \sqrt{3} \)[/tex].
### Part (b)
Equation: [tex]\(2 \log_4 (x-6) + \log_2 (x+6) = \frac{1}{\log_5 2} + \log_2 x\)[/tex]
1. Change of Base Formula: Rewrite all logarithms to base 2:
[tex]\[ \log_4 (x - 6) = \frac{\log_2 (x - 6)}{\log_2 4} = \frac{\log_2 (x - 6)}{2} \][/tex]
And use the fact that [tex]\(\frac{1}{\log_5 2} = \log_2 5\)[/tex]:
[tex]\[ 2 \cdot \frac{\log_2 (x - 6)}{2} + \log_2 (x + 6) = \log_2 5 + \log_2 x \][/tex]
2. Simplify:
[tex]\[ \log_2 (x - 6) + \log_2 (x + 6) = \log_2 5 + \log_2 x \][/tex]
3. Combine Logarithms:
[tex]\[ \log_2 \left[(x - 6)(x + 6)\right] = \log_2 (5x) \][/tex]
[tex]\[ \log_2 (x^2 - 36) = \log_2 (5x) \][/tex]
4. Set the Arguments Equal: Since the logarithmic functions are equal, their arguments must be equal too:
[tex]\[ x^2 - 36 = 5x \][/tex]
5. Solve the Quadratic Equation:
[tex]\[ x^2 - 5x - 36 = 0 \][/tex]
[tex]\[ (x - 9)(x + 4) = 0 \][/tex]
This gives us two potential solutions:
[tex]\[ x = 9 \quad \text{and} \quad x = -4 \][/tex]
6. Check for Validity: We discard [tex]\(x = -4\)[/tex] because the logarithms are not defined for negative arguments. Hence, the only valid solution is:
[tex]\[ x = 9 \][/tex]
Therefore, the solution to part (b) is [tex]\( x = 9 \)[/tex].
### Summary:
- The solutions to part (a) are [tex]\( x = 27 \)[/tex] and [tex]\( x = \sqrt{3} \)[/tex].
- The solution to part (b) is [tex]\( x = 9 \)[/tex].
### Part (a)
Equation: [tex]\(\log_x 27 + 2 \log_3 x = 7\)[/tex]
1. Change of Base Formula: Recall that [tex]\(\log_a b = \frac{\log b}{\log a}\)[/tex].
We can rewrite [tex]\(\log_x 27\)[/tex] and [tex]\(\log_3 x\)[/tex] using the change of base formula:
[tex]\[ \log_x 27 = \frac{\log 27}{\log x} \][/tex]
[tex]\[ \log_3 x = \frac{\log x}{\log 3} \][/tex]
2. Substitute these expressions into the equation:
[tex]\[ \frac{\log 27}{\log x} + 2 \cdot \frac{\log x}{\log 3} = 7 \][/tex]
3. Simplify: Combine terms over a common denominator.
[tex]\[ \frac{\log 27 \cdot \log 3 + 2 (\log x)^2}{\log x \cdot \log 3} = 7 \][/tex]
4. Log Value: Recall that [tex]\(27 = 3^3\)[/tex], thus [tex]\(\log 27 = \log (3^3) = 3 \log 3\)[/tex]. Substitute this into the equation:
[tex]\[ \frac{3 \log 3 \cdot \log 3 + 2 (\log x)^2}{\log x \cdot \log 3} = 7 \][/tex]
[tex]\[ \frac{3 (\log 3)^2 + 2 (\log x)^2}{\log x \cdot \log 3} = 7 \][/tex]
5. Combine and Rearrange to a standard quadratic form in terms of [tex]\(\log x\)[/tex]:
[tex]\[ 3 (\log 3)^2 + 2 (\log x)^2 = 7 \log x \cdot \log 3 \][/tex]
6. Solve the Quadratic Equation: Let [tex]\(y = \log x\)[/tex]. The equation becomes:
[tex]\[ 2y^2 - 7 (\log 3)y + 3 (\log 3)^2 = 0 \][/tex]
7. Factorize or Use the Quadratic Formula to find [tex]\(y\)[/tex] (i.e., [tex]\(\log x\)[/tex]):
[tex]\[ y = \frac{7 (\log 3) \pm \sqrt{(7 (\log 3))^2 - 4 \cdot 2 \cdot 3 (\log 3)^2}}{2 \cdot 2} \][/tex]
[tex]\[ y = \frac{7 \log 3 \pm \sqrt{49 (\log 3)^2 - 24 (\log 3)^2}}{4} \][/tex]
[tex]\[ y = \frac{7 \log 3 \pm 5 \log 3}{4} \][/tex]
This gives us two solutions:
[tex]\[ y = 3 (\log 3) \quad \text{and} \quad y = \frac{\log 3}{2} \][/tex]
8. Find [tex]\(x\)[/tex]:
[tex]\[ \log x = 3 (\log 3) \Rightarrow x = 3^3 = 27 \][/tex]
[tex]\[ \log x = \frac{\log 3}{2} \Rightarrow x = 3^{1/2} = \sqrt{3} \][/tex]
Thus, the solutions are [tex]\( x = 27 \)[/tex] and [tex]\( x = \sqrt{3} \)[/tex].
### Part (b)
Equation: [tex]\(2 \log_4 (x-6) + \log_2 (x+6) = \frac{1}{\log_5 2} + \log_2 x\)[/tex]
1. Change of Base Formula: Rewrite all logarithms to base 2:
[tex]\[ \log_4 (x - 6) = \frac{\log_2 (x - 6)}{\log_2 4} = \frac{\log_2 (x - 6)}{2} \][/tex]
And use the fact that [tex]\(\frac{1}{\log_5 2} = \log_2 5\)[/tex]:
[tex]\[ 2 \cdot \frac{\log_2 (x - 6)}{2} + \log_2 (x + 6) = \log_2 5 + \log_2 x \][/tex]
2. Simplify:
[tex]\[ \log_2 (x - 6) + \log_2 (x + 6) = \log_2 5 + \log_2 x \][/tex]
3. Combine Logarithms:
[tex]\[ \log_2 \left[(x - 6)(x + 6)\right] = \log_2 (5x) \][/tex]
[tex]\[ \log_2 (x^2 - 36) = \log_2 (5x) \][/tex]
4. Set the Arguments Equal: Since the logarithmic functions are equal, their arguments must be equal too:
[tex]\[ x^2 - 36 = 5x \][/tex]
5. Solve the Quadratic Equation:
[tex]\[ x^2 - 5x - 36 = 0 \][/tex]
[tex]\[ (x - 9)(x + 4) = 0 \][/tex]
This gives us two potential solutions:
[tex]\[ x = 9 \quad \text{and} \quad x = -4 \][/tex]
6. Check for Validity: We discard [tex]\(x = -4\)[/tex] because the logarithms are not defined for negative arguments. Hence, the only valid solution is:
[tex]\[ x = 9 \][/tex]
Therefore, the solution to part (b) is [tex]\( x = 9 \)[/tex].
### Summary:
- The solutions to part (a) are [tex]\( x = 27 \)[/tex] and [tex]\( x = \sqrt{3} \)[/tex].
- The solution to part (b) is [tex]\( x = 9 \)[/tex].
