Solve For { Y $}$ In The Equation:${ \left(\log _3 Y\right)^2 - \frac{1}{2} \log _3 Y = \frac{3}{2} }$

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Introduction

In this article, we will delve into the world of logarithmic equations and solve for y in the given equation: (log⁡3y)2−12log⁡3y=32\left(\log _3 y\right)^2 - \frac{1}{2} \log _3 y = \frac{3}{2}. This equation involves logarithms with base 3, and we will use various mathematical techniques to isolate y and find its value.

Understanding the Equation

The given equation is a quadratic equation in terms of log⁡3y\log _3 y. To solve for y, we need to first manipulate the equation to isolate log⁡3y\log _3 y. Let's start by rewriting the equation as follows:

(log⁡3y)2−12log⁡3y−32=0\left(\log _3 y\right)^2 - \frac{1}{2} \log _3 y - \frac{3}{2} = 0

This equation is a quadratic equation in terms of log⁡3y\log _3 y, and we can use the quadratic formula to solve for log⁡3y\log _3 y.

Using the Quadratic Formula

The quadratic formula states that for an equation of the form ax2+bx+c=0ax^2 + bx + c = 0, the solutions are given by:

x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

In our case, a=1a = 1, b=−12b = -\frac{1}{2}, and c=−32c = -\frac{3}{2}. Plugging these values into the quadratic formula, we get:

log⁡3y=12±(−12)2−4(1)(−32)2(1)\log _3 y = \frac{\frac{1}{2} \pm \sqrt{\left(-\frac{1}{2}\right)^2 - 4(1)\left(-\frac{3}{2}\right)}}{2(1)}

Simplifying the expression under the square root, we get:

log⁡3y=12±14+62\log _3 y = \frac{\frac{1}{2} \pm \sqrt{\frac{1}{4} + 6}}{2}

log⁡3y=12±2542\log _3 y = \frac{\frac{1}{2} \pm \sqrt{\frac{25}{4}}}{2}

log⁡3y=12±522\log _3 y = \frac{\frac{1}{2} \pm \frac{5}{2}}{2}

This gives us two possible values for log⁡3y\log _3 y:

log⁡3y=12+522=64=32\log _3 y = \frac{\frac{1}{2} + \frac{5}{2}}{2} = \frac{6}{4} = \frac{3}{2}

log⁡3y=12−522=−44=−1\log _3 y = \frac{\frac{1}{2} - \frac{5}{2}}{2} = \frac{-4}{4} = -1

Solving for y

Now that we have found the values of log⁡3y\log _3 y, we can solve for y by exponentiating both sides of the equation. Recall that log⁡3y=x\log _3 y = x is equivalent to y=3xy = 3^x. Therefore, we have:

y=332y = 3^{\frac{3}{2}}

y=3−1y = 3^{-1}

The first equation gives us y=33=27=33y = \sqrt{3^3} = \sqrt{27} = 3\sqrt{3}, and the second equation gives us y=13y = \frac{1}{3}.

Conclusion

In this article, we solved for y in the equation (log⁡3y)2−12log⁡3y=32\left(\log _3 y\right)^2 - \frac{1}{2} \log _3 y = \frac{3}{2}. We used the quadratic formula to find the values of log⁡3y\log _3 y, and then exponentiated both sides of the equation to solve for y. The two possible values for y are 333\sqrt{3} and 13\frac{1}{3}.

Additional Tips and Tricks

  • When solving logarithmic equations, it's often helpful to use the change-of-base formula to rewrite the equation in terms of a common logarithm.
  • The quadratic formula can be used to solve quadratic equations in terms of logarithms, but it's not the only method. Other methods, such as factoring or completing the square, may also be useful.
  • When solving for y in a logarithmic equation, it's essential to check the domain of the logarithm to ensure that the solution is valid.

Common Mistakes to Avoid

  • When using the quadratic formula, it's easy to make mistakes when simplifying the expression under the square root. Make sure to double-check your work to avoid errors.
  • When solving for y, it's essential to check the domain of the logarithm to ensure that the solution is valid. Failure to do so can lead to incorrect solutions.
  • When using the change-of-base formula, make sure to use the correct base for the logarithm. Using the wrong base can lead to incorrect solutions.

Real-World Applications

Logarithmic equations have many real-world applications in fields such as physics, engineering, and economics. For example, logarithmic equations can be used to model population growth, chemical reactions, and financial transactions.

Conclusion

Introduction

In our previous article, we solved for y in the equation (log⁡3y)2−12log⁡3y=32\left(\log _3 y\right)^2 - \frac{1}{2} \log _3 y = \frac{3}{2}. We used the quadratic formula to find the values of log⁡3y\log _3 y, and then exponentiated both sides of the equation to solve for y. In this article, we will answer some common questions that readers may have about solving logarithmic equations.

Q: What is the difference between a logarithmic equation and an exponential equation?

A: A logarithmic equation is an equation that involves a logarithm, which is the inverse of an exponential function. For example, the equation log⁡3y=2\log _3 y = 2 is a logarithmic equation, while the equation y=32y = 3^2 is an exponential equation.

Q: How do I know which base to use when solving a logarithmic equation?

A: The base of a logarithm is usually specified in the problem, but if it's not, you can use the change-of-base formula to rewrite the equation in terms of a common logarithm. For example, if you're given the equation log⁡5y=2\log _5 y = 2, you can rewrite it as log⁡3y=log⁡352\log _3 y = \log _3 5^2.

Q: Can I use the quadratic formula to solve any logarithmic equation?

A: Yes, the quadratic formula can be used to solve any logarithmic equation that can be rewritten in the form (log⁡by)2+clog⁡by+d=0\left(\log _b y\right)^2 + c \log _b y + d = 0. However, if the equation cannot be rewritten in this form, you may need to use other methods, such as factoring or completing the square.

Q: How do I check the domain of a logarithm to ensure that the solution is valid?

A: To check the domain of a logarithm, you need to make sure that the argument of the logarithm (the value inside the logarithm) is positive. For example, if you're given the equation log⁡3y=2\log _3 y = 2, you need to make sure that y>0y > 0.

Q: Can I use a calculator to solve a logarithmic equation?

A: Yes, you can use a calculator to solve a logarithmic equation, but you need to make sure that you're using the correct base and that the calculator is set to the correct mode (e.g. logarithmic mode).

Q: How do I know which method to use when solving a logarithmic equation?

A: The method you use to solve a logarithmic equation will depend on the form of the equation and the values of the coefficients. If the equation can be rewritten in the form (log⁡by)2+clog⁡by+d=0\left(\log _b y\right)^2 + c \log _b y + d = 0, you can use the quadratic formula. If the equation cannot be rewritten in this form, you may need to use other methods, such as factoring or completing the square.

Q: Can I use logarithmic equations to model real-world phenomena?

A: Yes, logarithmic equations can be used to model a wide range of real-world phenomena, including population growth, chemical reactions, and financial transactions.

Q: How do I apply logarithmic equations to real-world problems?

A: To apply logarithmic equations to real-world problems, you need to identify the variables and the relationships between them. You can then use the logarithmic equation to model the problem and solve for the unknown variable.

Conclusion

In this article, we answered some common questions that readers may have about solving logarithmic equations. We discussed the difference between logarithmic and exponential equations, how to choose the base of a logarithm, and how to check the domain of a logarithm. We also discussed how to use the quadratic formula to solve logarithmic equations and how to apply logarithmic equations to real-world problems.