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Sagot :
Let's solve the given logarithmic equations step by step.
### Part (a):
Given the equation:
[tex]\[ \log_2 x + \log_4 x = 4 \frac{1}{2} \][/tex]
First, let's express [tex]\(\log_4 x\)[/tex] in terms of [tex]\(\log_2 x\)[/tex]:
[tex]\[ \log_4 x = \log_{2^2} x = \frac{1}{2} \log_2 x \][/tex]
Substituting this back into the original equation, we get:
[tex]\[ \log_2 x + \frac{1}{2} \log_2 x = 4.5 \][/tex]
Combining the terms on the left-hand side:
[tex]\[ \frac{3}{2} \log_2 x = 4.5 \][/tex]
To isolate [tex]\(\log_2 x\)[/tex], multiply both sides by [tex]\(\frac{2}{3}\)[/tex]:
[tex]\[ \log_2 x = 4.5 \times \frac{2}{3} = 3 \][/tex]
Now, convert this logarithmic form to its exponential form:
[tex]\[ x = 2^3 = 8 \][/tex]
Thus, the solution to Part (a) is:
[tex]\[ x = 8 \][/tex]
### Part (b):
Given the equation:
[tex]\[ \log(x+2) + \log(x-1) = 1 \][/tex]
First, combine the logarithms on the left-hand side:
[tex]\[ \log[(x+2)(x-1)] = 1 \][/tex]
Convert this logarithmic equation into its exponential form:
[tex]\[ (x+2)(x-1) = 10 \][/tex]
Expanding and simplifying the quadratic equation:
[tex]\[ x^2 + x - 2 = 10 \][/tex]
[tex]\[ x^2 + x - 12 = 0 \][/tex]
Now solve this quadratic equation using the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 1\)[/tex], [tex]\(b = 1\)[/tex], and [tex]\(c = -12\)[/tex]:
Calculate the discriminant:
[tex]\[ b^2 - 4ac = 1^2 - 4(1)(-12) = 1 + 48 = 49 \][/tex]
So, the solutions are:
[tex]\[ x = \frac{-1 \pm \sqrt{49}}{2} = \frac{-1 \pm 7}{2} \][/tex]
This yields two possible solutions:
[tex]\[ x = \frac{-1 + 7}{2} = 3 \quad \text{and} \quad x = \frac{-1 - 7}{2} = -4 \][/tex]
Among the two solutions, [tex]\( x = -4 \)[/tex] is not valid because it would result in the logarithm of a negative number, which is undefined in the real number system. Therefore, the valid solution is:
[tex]\[ x = 3 \][/tex]
Thus, the solution to Part (b) is:
[tex]\[ x = 3 \][/tex]
### Summary:
- For part (a), the solution is [tex]\(x = 8\)[/tex].
- For part (b), the solution is [tex]\(x = 3\)[/tex].
### Part (a):
Given the equation:
[tex]\[ \log_2 x + \log_4 x = 4 \frac{1}{2} \][/tex]
First, let's express [tex]\(\log_4 x\)[/tex] in terms of [tex]\(\log_2 x\)[/tex]:
[tex]\[ \log_4 x = \log_{2^2} x = \frac{1}{2} \log_2 x \][/tex]
Substituting this back into the original equation, we get:
[tex]\[ \log_2 x + \frac{1}{2} \log_2 x = 4.5 \][/tex]
Combining the terms on the left-hand side:
[tex]\[ \frac{3}{2} \log_2 x = 4.5 \][/tex]
To isolate [tex]\(\log_2 x\)[/tex], multiply both sides by [tex]\(\frac{2}{3}\)[/tex]:
[tex]\[ \log_2 x = 4.5 \times \frac{2}{3} = 3 \][/tex]
Now, convert this logarithmic form to its exponential form:
[tex]\[ x = 2^3 = 8 \][/tex]
Thus, the solution to Part (a) is:
[tex]\[ x = 8 \][/tex]
### Part (b):
Given the equation:
[tex]\[ \log(x+2) + \log(x-1) = 1 \][/tex]
First, combine the logarithms on the left-hand side:
[tex]\[ \log[(x+2)(x-1)] = 1 \][/tex]
Convert this logarithmic equation into its exponential form:
[tex]\[ (x+2)(x-1) = 10 \][/tex]
Expanding and simplifying the quadratic equation:
[tex]\[ x^2 + x - 2 = 10 \][/tex]
[tex]\[ x^2 + x - 12 = 0 \][/tex]
Now solve this quadratic equation using the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 1\)[/tex], [tex]\(b = 1\)[/tex], and [tex]\(c = -12\)[/tex]:
Calculate the discriminant:
[tex]\[ b^2 - 4ac = 1^2 - 4(1)(-12) = 1 + 48 = 49 \][/tex]
So, the solutions are:
[tex]\[ x = \frac{-1 \pm \sqrt{49}}{2} = \frac{-1 \pm 7}{2} \][/tex]
This yields two possible solutions:
[tex]\[ x = \frac{-1 + 7}{2} = 3 \quad \text{and} \quad x = \frac{-1 - 7}{2} = -4 \][/tex]
Among the two solutions, [tex]\( x = -4 \)[/tex] is not valid because it would result in the logarithm of a negative number, which is undefined in the real number system. Therefore, the valid solution is:
[tex]\[ x = 3 \][/tex]
Thus, the solution to Part (b) is:
[tex]\[ x = 3 \][/tex]
### Summary:
- For part (a), the solution is [tex]\(x = 8\)[/tex].
- For part (b), the solution is [tex]\(x = 3\)[/tex].
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