IDNLearn.com: Your reliable source for finding precise answers. Ask anything and receive immediate, well-informed answers from our dedicated community of experts.
Sagot :
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].
We value your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. Your questions deserve precise answers. Thank you for visiting IDNLearn.com, and see you again soon for more helpful information.
