Solved on Nov 18, 2023

Find the value of $x$ that satisfies the equation $\ln(x-3) + \ln(3x+1) = 8$ .

STEP 1

Assumptions1. The equation is $\ln (x-3)+\ln (3x+1)=8$ . We are solving for $x$
3. The logarithm is a natural logarithm, meaning it is base $e$

STEP 2

We can use the property of logarithms that states the sum of two logarithms is equal to the logarithm of their product. So, we can combine the two logarithms on the left side of the equation into one.
$\ln (x-)+\ln (x+1)=\ln((x-)(x+1))$

STEP 3

Now, we can rewrite the equation as follows $\ln((x-3)(3x+1))=8$

STEP 4

To remove the logarithm, we can use the property of logarithms that states $a=\ln b$ is equivalent to $e^a=b$ . In this case, we can rewrite the equation as follows $(x-3)(3x+1)=e^8$

STEP 5

Now, we can simplify the equation by expanding the product on the left side of the equation.
$3x^2 -8x -3 = e^8$

STEP 6

We now have a quadratic equation. To solve for $x$ , we can use the quadratic formula. But first, we need to move $e^8$ to the left side of the equation to set the equation equal to zero.
$3x^2 -8x -3 - e^8 =0$

STEP 7

Now, we can use the quadratic formula to solve for $x$ . The quadratic formula is $x = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}$ where $a$ , $b$ , and $c$ are the coefficients of the quadratic equation.

STEP 8

In our equation, $a =3$ , $b = -8$ , and $c = -3 - e^8$ . Plugging these values into the quadratic formula gives us $x = \frac{-(-8) \pm \sqrt{(-8)^2 -43(-3 - e^8)}}{2*3}$

STEP 9

olving the above equation will give us the values of $x$ . However, we need to remember that the domain of the original equation (due to the logarithms) is $x >3$ and $x > -/3$ , so we need to check if the solutions satisfy these conditions.

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Find the value of $x$ that satisfies the equation $\ln(x-3) + \ln(3x+1) = 8$ .

STEP 1

Assumptions1. The equation is $\ln (x-3)+\ln (3x+1)=8$ . We are solving for $x$
3. The logarithm is a natural logarithm, meaning it is base $e$

STEP 2

We can use the property of logarithms that states the sum of two logarithms is equal to the logarithm of their product. So, we can combine the two logarithms on the left side of the equation into one.
$\ln (x-)+\ln (x+1)=\ln((x-)(x+1))$

STEP 3

Now, we can rewrite the equation as follows $\ln((x-3)(3x+1))=8$

STEP 4

To remove the logarithm, we can use the property of logarithms that states $a=\ln b$ is equivalent to $e^a=b$ . In this case, we can rewrite the equation as follows $(x-3)(3x+1)=e^8$

STEP 5

Now, we can simplify the equation by expanding the product on the left side of the equation.
$3x^2 -8x -3 = e^8$

STEP 6

We now have a quadratic equation. To solve for $x$ , we can use the quadratic formula. But first, we need to move $e^8$ to the left side of the equation to set the equation equal to zero.
$3x^2 -8x -3 - e^8 =0$

STEP 7

Now, we can use the quadratic formula to solve for $x$ . The quadratic formula is $x = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}$ where $a$ , $b$ , and $c$ are the coefficients of the quadratic equation.

STEP 8

In our equation, $a =3$ , $b = -8$ , and $c = -3 - e^8$ . Plugging these values into the quadratic formula gives us $x = \frac{-(-8) \pm \sqrt{(-8)^2 -43(-3 - e^8)}}{2*3}$

STEP 9

olving the above equation will give us the values of $x$ . However, we need to remember that the domain of the original equation (due to the logarithms) is $x >3$ and $x > -/3$ , so we need to check if the solutions satisfy these conditions.

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Find the value of xxx that satisfies the equation ln⁡(x−3)+ln⁡(3x+1)=8\ln(x-3) + \ln(3x+1) = 8ln(x−3)+ln(3x+1)=8.

STEP 1

Assumptions1. The equation is ln⁡(x−3)+ln⁡(3x+1)=8\ln (x-3)+\ln (3x+1)=8ln(x−3)+ln(3x+1)=8 . We are solving for xxx3. The logarithm is a natural logarithm, meaning it is base eee

STEP 2

STEP 3

Now, we can rewrite the equation as followsln⁡((x−3)(3x+1))=8\ln((x-3)(3x+1))=8ln((x−3)(3x+1))=8

STEP 4

To remove the logarithm, we can use the property of logarithms that states a=ln⁡ba=\ln ba=lnb is equivalent to ea=be^a=bea=b. In this case, we can rewrite the equation as follows(x−3)(3x+1)=e8(x-3)(3x+1)=e^8(x−3)(3x+1)=e8

STEP 5

Now, we can simplify the equation by expanding the product on the left side of the equation.3x2−8x−3=e83x^2 -8x -3 = e^83x2−8x−3=e8

STEP 6

We now have a quadratic equation. To solve for xxx, we can use the quadratic formula. But first, we need to move e8e^8e8 to the left side of the equation to set the equation equal to zero.3x2−8x−3−e8=03x^2 -8x -3 - e^8 =03x2−8x−3−e8=0

STEP 7

Now, we can use the quadratic formula to solve for xxx. The quadratic formula isx=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}x=2a−b±b2−4ac​​where aaa, bbb, and ccc are the coefficients of the quadratic equation.

STEP 8

STEP 9

olving the above equation will give us the values of xxx. However, we need to remember that the domain of the original equation (due to the logarithms) is x>3x >3x>3 and x>−/3x > -/3x>−/3, so we need to check if the solutions satisfy these conditions.

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Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Find the value of $x$ that satisfies the equation $\ln(x-3) + \ln(3x+1) = 8$ .

Assumptions1. The equation is $\ln (x-3)+\ln (3x+1)=8$ . We are solving for $x$
3. The logarithm is a natural logarithm, meaning it is base $e$

Now, we can rewrite the equation as follows $\ln((x-3)(3x+1))=8$

To remove the logarithm, we can use the property of logarithms that states $a=\ln b$ is equivalent to $e^a=b$ . In this case, we can rewrite the equation as follows $(x-3)(3x+1)=e^8$

Now, we can simplify the equation by expanding the product on the left side of the equation.
$3x^2 -8x -3 = e^8$

We now have a quadratic equation. To solve for $x$ , we can use the quadratic formula. But first, we need to move $e^8$ to the left side of the equation to set the equation equal to zero.
$3x^2 -8x -3 - e^8 =0$

Now, we can use the quadratic formula to solve for $x$ . The quadratic formula is $x = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}$ where $a$ , $b$ , and $c$ are the coefficients of the quadratic equation.

olving the above equation will give us the values of $x$ . However, we need to remember that the domain of the original equation (due to the logarithms) is $x >3$ and $x > -/3$ , so we need to check if the solutions satisfy these conditions.