Answer :
To solve the equation [tex]\( 2 \log_5(3x - 2) - \log_5(x) = 2 \)[/tex], follow these steps:
1. Express the equation in terms of logarithms:
We start with the given equation:
[tex]\[ 2 \log_5(3x - 2) - \log_5(x) = 2 \][/tex]
2. Combine the logarithmic terms:
Recall the properties of logarithms:
[tex]\[ a \log_b(c) = \log_b(c^a) \][/tex]
and
[tex]\[ \log_b(a) - \log_b(b) = \log_b\left(\frac{a}{b}\right) \][/tex]
So we use the first property to rewrite [tex]\(2 \log_5(3x - 2)\)[/tex] as:
[tex]\[ 2 \log_5(3x - 2) = \log_5((3x - 2)^2) \][/tex]
Then our equation becomes:
[tex]\[ \log_5((3x - 2)^2) - \log_5(x) = 2 \][/tex]
Using the second property of logarithms, combine the logarithmic terms:
[tex]\[ \log_5\left(\frac{(3x - 2)^2}{x}\right) = 2 \][/tex]
3. Rewrite the equation in exponential form:
The equation [tex]\(\log_b(A) = C\)[/tex] is equivalent to [tex]\(A = b^C\)[/tex].
Thus, we can rewrite [tex]\(\log_5\left(\frac{(3x - 2)^2}{x}\right) = 2\)[/tex] as:
[tex]\[ \frac{(3x - 2)^2}{x} = 5^2 \][/tex]
Simplify this:
[tex]\[ \frac{(3x - 2)^2}{x} = 25 \][/tex]
4. Solve the resulting equation:
Multiply both sides by [tex]\(x\)[/tex] to eliminate the fraction:
[tex]\[ (3x - 2)^2 = 25x \][/tex]
Expand the left-hand side:
[tex]\[ 9x^2 - 12x + 4 = 25x \][/tex]
Bring all terms to one side to set the equation to zero:
[tex]\[ 9x^2 - 12x + 4 - 25x = 0 \][/tex]
[tex]\[ 9x^2 - 37x + 4 = 0 \][/tex]
5. Solve the quadratic equation:
Use the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 9\)[/tex], [tex]\(b = -37\)[/tex], and [tex]\(c = 4\)[/tex]:
[tex]\[ x = \frac{-(-37) \pm \sqrt{(-37)^2 - 4 \cdot 9 \cdot 4}}{2 \cdot 9} \][/tex]
[tex]\[ x = \frac{37 \pm \sqrt{1369 - 144}}{18} \][/tex]
[tex]\[ x = \frac{37 \pm \sqrt{1225}}{18} \][/tex]
[tex]\[ x = \frac{37 \pm 35}{18} \][/tex]
This gives us two possible solutions:
[tex]\[ x = \frac{37 + 35}{18} = \frac{72}{18} = 4 \][/tex]
[tex]\[ x = \frac{37 - 35}{18} = \frac{2}{18} = \frac{1}{9} \][/tex]
6. Verify the solutions:
Substitute [tex]\(x = 4\)[/tex] and [tex]\(x = \frac{1}{9}\)[/tex] back into the original equation to check for validity:
- For [tex]\(x = 4\)[/tex]:
[tex]\[ 2 \log_5(3 \cdot 4 - 2) - \log_5(4) = 2 \][/tex]
[tex]\[ 2 \log_5(10) - \log_5(4) = 2 \][/tex]
[tex]\[ 2 \log_5(10) - \log_5(4) = 2 \][/tex]
Simplifying the left-hand side confirms that the equation holds true.
- For [tex]\(x = \frac{1}{9}\)[/tex]:
Without calculation, substitute [tex]\(x = \frac{1}{9}\)[/tex] into the equation results negative arguments in the logarithms which are undefined in the real number ([3x - 2 < 0]). Therefore, this solution doesn't satisfy the original equation,
Thus, the valid solution is:
[tex]\[ x = 4 \][/tex]
1. Express the equation in terms of logarithms:
We start with the given equation:
[tex]\[ 2 \log_5(3x - 2) - \log_5(x) = 2 \][/tex]
2. Combine the logarithmic terms:
Recall the properties of logarithms:
[tex]\[ a \log_b(c) = \log_b(c^a) \][/tex]
and
[tex]\[ \log_b(a) - \log_b(b) = \log_b\left(\frac{a}{b}\right) \][/tex]
So we use the first property to rewrite [tex]\(2 \log_5(3x - 2)\)[/tex] as:
[tex]\[ 2 \log_5(3x - 2) = \log_5((3x - 2)^2) \][/tex]
Then our equation becomes:
[tex]\[ \log_5((3x - 2)^2) - \log_5(x) = 2 \][/tex]
Using the second property of logarithms, combine the logarithmic terms:
[tex]\[ \log_5\left(\frac{(3x - 2)^2}{x}\right) = 2 \][/tex]
3. Rewrite the equation in exponential form:
The equation [tex]\(\log_b(A) = C\)[/tex] is equivalent to [tex]\(A = b^C\)[/tex].
Thus, we can rewrite [tex]\(\log_5\left(\frac{(3x - 2)^2}{x}\right) = 2\)[/tex] as:
[tex]\[ \frac{(3x - 2)^2}{x} = 5^2 \][/tex]
Simplify this:
[tex]\[ \frac{(3x - 2)^2}{x} = 25 \][/tex]
4. Solve the resulting equation:
Multiply both sides by [tex]\(x\)[/tex] to eliminate the fraction:
[tex]\[ (3x - 2)^2 = 25x \][/tex]
Expand the left-hand side:
[tex]\[ 9x^2 - 12x + 4 = 25x \][/tex]
Bring all terms to one side to set the equation to zero:
[tex]\[ 9x^2 - 12x + 4 - 25x = 0 \][/tex]
[tex]\[ 9x^2 - 37x + 4 = 0 \][/tex]
5. Solve the quadratic equation:
Use the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 9\)[/tex], [tex]\(b = -37\)[/tex], and [tex]\(c = 4\)[/tex]:
[tex]\[ x = \frac{-(-37) \pm \sqrt{(-37)^2 - 4 \cdot 9 \cdot 4}}{2 \cdot 9} \][/tex]
[tex]\[ x = \frac{37 \pm \sqrt{1369 - 144}}{18} \][/tex]
[tex]\[ x = \frac{37 \pm \sqrt{1225}}{18} \][/tex]
[tex]\[ x = \frac{37 \pm 35}{18} \][/tex]
This gives us two possible solutions:
[tex]\[ x = \frac{37 + 35}{18} = \frac{72}{18} = 4 \][/tex]
[tex]\[ x = \frac{37 - 35}{18} = \frac{2}{18} = \frac{1}{9} \][/tex]
6. Verify the solutions:
Substitute [tex]\(x = 4\)[/tex] and [tex]\(x = \frac{1}{9}\)[/tex] back into the original equation to check for validity:
- For [tex]\(x = 4\)[/tex]:
[tex]\[ 2 \log_5(3 \cdot 4 - 2) - \log_5(4) = 2 \][/tex]
[tex]\[ 2 \log_5(10) - \log_5(4) = 2 \][/tex]
[tex]\[ 2 \log_5(10) - \log_5(4) = 2 \][/tex]
Simplifying the left-hand side confirms that the equation holds true.
- For [tex]\(x = \frac{1}{9}\)[/tex]:
Without calculation, substitute [tex]\(x = \frac{1}{9}\)[/tex] into the equation results negative arguments in the logarithms which are undefined in the real number ([3x - 2 < 0]). Therefore, this solution doesn't satisfy the original equation,
Thus, the valid solution is:
[tex]\[ x = 4 \][/tex]
