Answer :
To solve the equation [tex]\(\log_x(4x - 3) = 2\)[/tex], we will proceed through a step-by-step solution:
1. Understand the logarithmic form:
[tex]\[ \log_x(4x - 3) = 2 \][/tex]
This equation states that [tex]\( x \)[/tex] raised to the power of 2 is equal to [tex]\( 4x - 3 \)[/tex].
2. Convert the logarithmic equation to its exponential form:
[tex]\[ x^2 = 4x - 3 \][/tex]
3. Rearrange the equation to form a standard quadratic equation:
[tex]\[ x^2 - 4x + 3 = 0 \][/tex]
4. Solve the quadratic equation:
The quadratic equation [tex]\( x^2 - 4x + 3 = 0 \)[/tex] can be solved using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, [tex]\( a = 1 \)[/tex], [tex]\( b = -4 \)[/tex], and [tex]\( c = 3 \)[/tex].
5. Calculate the discriminant:
[tex]\[ \text{Discriminant} = b^2 - 4ac = (-4)^2 - 4(1)(3) = 16 - 12 = 4 \][/tex]
6. Find the roots using the quadratic formula:
[tex]\[ x = \frac{-(-4) \pm \sqrt{4}}{2(1)} = \frac{4 \pm 2}{2} \][/tex]
This gives us two solutions:
[tex]\[ x_1 = \frac{4 + 2}{2} = 3 \][/tex]
[tex]\[ x_2 = \frac{4 - 2}{2} = 1 \][/tex]
7. Determine the valid root(s) based on the constraints for the logarithm:
Since the logarithm [tex]\(\log_x(4x - 3)\)[/tex] is defined only if [tex]\(x > 1\)[/tex] and [tex]\(4x - 3 > 0\)[/tex]:
- For [tex]\(4x - 3 > 0\)[/tex]:
[tex]\[ 4x - 3 > 0 \implies x > \frac{3}{4} \][/tex]
- Therefore, [tex]\(x\)[/tex] must satisfy both [tex]\(x > \frac{3}{4}\)[/tex] and [tex]\(x > 1\)[/tex].
8. Check the solutions:
- [tex]\( x_1 = 3 \)[/tex]: This is valid since [tex]\(3 > 1\)[/tex].
- [tex]\( x_2 = 1 \)[/tex]: This is not valid since [tex]\(1\)[/tex] is not greater than [tex]\(1\)[/tex].
Hence, the only valid solution is:
[tex]\[ x = 3 \][/tex]
Therefore, the value of [tex]\( x \)[/tex] for which the logarithm is defined and the equation [tex]\(\log_x(4x - 3) = 2\)[/tex] holds true is:
[tex]\[ \boxed{3} \][/tex]
So, the correct answer is:
a. 3
1. Understand the logarithmic form:
[tex]\[ \log_x(4x - 3) = 2 \][/tex]
This equation states that [tex]\( x \)[/tex] raised to the power of 2 is equal to [tex]\( 4x - 3 \)[/tex].
2. Convert the logarithmic equation to its exponential form:
[tex]\[ x^2 = 4x - 3 \][/tex]
3. Rearrange the equation to form a standard quadratic equation:
[tex]\[ x^2 - 4x + 3 = 0 \][/tex]
4. Solve the quadratic equation:
The quadratic equation [tex]\( x^2 - 4x + 3 = 0 \)[/tex] can be solved using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, [tex]\( a = 1 \)[/tex], [tex]\( b = -4 \)[/tex], and [tex]\( c = 3 \)[/tex].
5. Calculate the discriminant:
[tex]\[ \text{Discriminant} = b^2 - 4ac = (-4)^2 - 4(1)(3) = 16 - 12 = 4 \][/tex]
6. Find the roots using the quadratic formula:
[tex]\[ x = \frac{-(-4) \pm \sqrt{4}}{2(1)} = \frac{4 \pm 2}{2} \][/tex]
This gives us two solutions:
[tex]\[ x_1 = \frac{4 + 2}{2} = 3 \][/tex]
[tex]\[ x_2 = \frac{4 - 2}{2} = 1 \][/tex]
7. Determine the valid root(s) based on the constraints for the logarithm:
Since the logarithm [tex]\(\log_x(4x - 3)\)[/tex] is defined only if [tex]\(x > 1\)[/tex] and [tex]\(4x - 3 > 0\)[/tex]:
- For [tex]\(4x - 3 > 0\)[/tex]:
[tex]\[ 4x - 3 > 0 \implies x > \frac{3}{4} \][/tex]
- Therefore, [tex]\(x\)[/tex] must satisfy both [tex]\(x > \frac{3}{4}\)[/tex] and [tex]\(x > 1\)[/tex].
8. Check the solutions:
- [tex]\( x_1 = 3 \)[/tex]: This is valid since [tex]\(3 > 1\)[/tex].
- [tex]\( x_2 = 1 \)[/tex]: This is not valid since [tex]\(1\)[/tex] is not greater than [tex]\(1\)[/tex].
Hence, the only valid solution is:
[tex]\[ x = 3 \][/tex]
Therefore, the value of [tex]\( x \)[/tex] for which the logarithm is defined and the equation [tex]\(\log_x(4x - 3) = 2\)[/tex] holds true is:
[tex]\[ \boxed{3} \][/tex]
So, the correct answer is:
a. 3
