Answer :
To show that [tex]\(\log_{\frac{1}{a}} y = -\log_a y\)[/tex], we start with the definition of logarithms. By definition, if [tex]\(\log_b(y) = x\)[/tex], then [tex]\(b^x = y\)[/tex].
For the logarithm with the base [tex]\(\frac{1}{a}\)[/tex], we write:
[tex]\[ \log_{\frac{1}{a}} y = x \implies \left(\frac{1}{a}\right)^x = y \][/tex]
We know that [tex]\(\frac{1}{a} = a^{-1}\)[/tex], so we can rewrite the left-hand side:
[tex]\[ a^{-x} = y \][/tex]
By definition of the logarithm with base [tex]\(a\)[/tex], if [tex]\(a^z = y\)[/tex], then [tex]\(z = \log_a y\)[/tex]. Here, we have:
[tex]\[ a^{-x} = y \implies -x = \log_a y \implies x = -\log_a y \][/tex]
Therefore:
[tex]\[ \log_{\frac{1}{a}} y = -\log_a y \][/tex]
This property is now established.
Next, we use this property to solve the equation:
[tex]\[ \log_3(3 - x) = \log_{\frac{1}{3}}(5 - 2x) \][/tex]
Using the identity we just proved, we can rewrite the right-hand side:
[tex]\[ \log_3(3 - x) = -\log_3(5 - 2x) \][/tex]
Rewriting using the negative sign:
[tex]\[ \log_3(3 - x) = \log_3((5 - 2x)^{-1}) \][/tex]
Since the logarithms are equal, we can equate the arguments:
[tex]\[ 3 - x = (5 - 2x)^{-1} \][/tex]
Taking the reciprocal on both sides gives:
[tex]\[ (3 - x)^{-1} = 5 - 2x \][/tex]
Now, solving for [tex]\(x\)[/tex]:
1. Start with:
[tex]\[ \frac{1}{3 - x} = 5 - 2x \][/tex]
2. Cross-multiply to clear the fraction:
[tex]\[ 1 = (3 - x)(5 - 2x) \][/tex]
3. Expand the right-hand side:
[tex]\[ 1 = 15 - 6x - 5x + 2x^2 \][/tex]
[tex]\[ 1 = 15 - 11x + 2x^2 \][/tex]
4. Rearrange into standard quadratic form:
[tex]\[ 2x^2 - 11x + 15 - 1 = 0 \][/tex]
[tex]\[ 2x^2 - 11x + 14 = 0 \][/tex]
5. Solve this quadratic equation using the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 2\)[/tex], [tex]\(b = -11\)[/tex], and [tex]\(c = 14\)[/tex]:
[tex]\[ x = \frac{11 \pm \sqrt{(-11)^2 - 4 \cdot 2 \cdot 14}}{2 \cdot 2} \][/tex]
[tex]\[ x = \frac{11 \pm \sqrt{121 - 112}}{4} \][/tex]
[tex]\[ x = \frac{11 \pm \sqrt{9}}{4} \][/tex]
[tex]\[ x = \frac{11 \pm 3}{4} \][/tex]
6. This gives two possible solutions:
[tex]\[ x = \frac{11 + 3}{4} = \frac{14}{4} = 3.5 \][/tex]
[tex]\[ x = \frac{11 - 3}{4} = \frac{8}{4} = 2 \][/tex]
Among these solutions, we need to verify which is valid for the original equation. Checking the valid solution:
For [tex]\(x = 3.5\)[/tex]:
[tex]\[ 3 - 3.5 = -0.5 \quad (\text{which does not work as log of a negative number is undefined}) \][/tex]
For [tex]\(x = 2\)[/tex]:
[tex]\[ 3 - 2 = 1 \quad \text{and} \quad 5 - 2(2) = 1 \quad (\text{valid}) \][/tex]
Thus, the valid solution is:
[tex]\[ x = 2 \][/tex]
For the logarithm with the base [tex]\(\frac{1}{a}\)[/tex], we write:
[tex]\[ \log_{\frac{1}{a}} y = x \implies \left(\frac{1}{a}\right)^x = y \][/tex]
We know that [tex]\(\frac{1}{a} = a^{-1}\)[/tex], so we can rewrite the left-hand side:
[tex]\[ a^{-x} = y \][/tex]
By definition of the logarithm with base [tex]\(a\)[/tex], if [tex]\(a^z = y\)[/tex], then [tex]\(z = \log_a y\)[/tex]. Here, we have:
[tex]\[ a^{-x} = y \implies -x = \log_a y \implies x = -\log_a y \][/tex]
Therefore:
[tex]\[ \log_{\frac{1}{a}} y = -\log_a y \][/tex]
This property is now established.
Next, we use this property to solve the equation:
[tex]\[ \log_3(3 - x) = \log_{\frac{1}{3}}(5 - 2x) \][/tex]
Using the identity we just proved, we can rewrite the right-hand side:
[tex]\[ \log_3(3 - x) = -\log_3(5 - 2x) \][/tex]
Rewriting using the negative sign:
[tex]\[ \log_3(3 - x) = \log_3((5 - 2x)^{-1}) \][/tex]
Since the logarithms are equal, we can equate the arguments:
[tex]\[ 3 - x = (5 - 2x)^{-1} \][/tex]
Taking the reciprocal on both sides gives:
[tex]\[ (3 - x)^{-1} = 5 - 2x \][/tex]
Now, solving for [tex]\(x\)[/tex]:
1. Start with:
[tex]\[ \frac{1}{3 - x} = 5 - 2x \][/tex]
2. Cross-multiply to clear the fraction:
[tex]\[ 1 = (3 - x)(5 - 2x) \][/tex]
3. Expand the right-hand side:
[tex]\[ 1 = 15 - 6x - 5x + 2x^2 \][/tex]
[tex]\[ 1 = 15 - 11x + 2x^2 \][/tex]
4. Rearrange into standard quadratic form:
[tex]\[ 2x^2 - 11x + 15 - 1 = 0 \][/tex]
[tex]\[ 2x^2 - 11x + 14 = 0 \][/tex]
5. Solve this quadratic equation using the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 2\)[/tex], [tex]\(b = -11\)[/tex], and [tex]\(c = 14\)[/tex]:
[tex]\[ x = \frac{11 \pm \sqrt{(-11)^2 - 4 \cdot 2 \cdot 14}}{2 \cdot 2} \][/tex]
[tex]\[ x = \frac{11 \pm \sqrt{121 - 112}}{4} \][/tex]
[tex]\[ x = \frac{11 \pm \sqrt{9}}{4} \][/tex]
[tex]\[ x = \frac{11 \pm 3}{4} \][/tex]
6. This gives two possible solutions:
[tex]\[ x = \frac{11 + 3}{4} = \frac{14}{4} = 3.5 \][/tex]
[tex]\[ x = \frac{11 - 3}{4} = \frac{8}{4} = 2 \][/tex]
Among these solutions, we need to verify which is valid for the original equation. Checking the valid solution:
For [tex]\(x = 3.5\)[/tex]:
[tex]\[ 3 - 3.5 = -0.5 \quad (\text{which does not work as log of a negative number is undefined}) \][/tex]
For [tex]\(x = 2\)[/tex]:
[tex]\[ 3 - 2 = 1 \quad \text{and} \quad 5 - 2(2) = 1 \quad (\text{valid}) \][/tex]
Thus, the valid solution is:
[tex]\[ x = 2 \][/tex]
