Solving Logarithmic Equations: A Step-by-Step Guide

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Solving Logarithmic Equations and Evaluating Logarithms

Hey guys! Let's dive into some math problems today. We're going to break down how to solve logarithmic equations and evaluate logarithms. Don't worry, it might seem tricky at first, but with a little practice, you'll be acing these in no time. We'll go through each problem step by step, making sure everything is super clear and easy to understand. Ready to get started?

Solving the Equation: b. 133log⁑3x=2\frac{1}{3} \sqrt{3} \log_{3x} = 2

Alright, let's tackle this logarithmic equation together. The goal here is to isolate x. This might seem a bit daunting at first, but we'll take it slow and break it down. Our main aim is to get that x out of the logarithm so we can actually solve for it. So, let's start by trying to isolate the logarithmic term on one side of the equation. We have: 133log⁑3x=2\frac{1}{3} \sqrt{3} \log_{3x} = 2. First, we want to get rid of the coefficient in front of the logarithm. This is 133\frac{1}{3} \sqrt{3}.

To do this, we'll multiply both sides of the equation by the reciprocal of this coefficient. The reciprocal of 133\frac{1}{3} \sqrt{3} is 33\frac{3}{\sqrt{3}}.

So, let's multiply both sides by 33\frac{3}{\sqrt{3}}:

33βˆ—133log⁑3x=2βˆ—33\frac{3}{\sqrt{3}} * \frac{1}{3} \sqrt{3} \log_{3x} = 2 * \frac{3}{\sqrt{3}}

This simplifies to: log⁑3x=63\log_{3x} = \frac{6}{\sqrt{3}}.

Now, let's rationalize the denominator on the right side. We multiply both the numerator and denominator by 3\sqrt{3}:

63βˆ—33=633=23\frac{6}{\sqrt{3}} * \frac{\sqrt{3}}{\sqrt{3}} = \frac{6\sqrt{3}}{3} = 2\sqrt{3}.

So, our equation is now: log⁑3x=23\log_{3x} = 2\sqrt{3}.

Next, we'll convert the logarithmic form to exponential form. Remember that log⁑ba=c\log_{b} a = c is equivalent to bc=ab^c = a. Applying this to our equation, log⁑3x=23\log_{3x} = 2\sqrt{3}, we get: (3x)23=10(3x)^{2\sqrt{3}} = 10. Oops! I made a mistake here, the original equation is 13βˆ—log3x=2\frac{1}{\sqrt{3}}*log_{3x} = 2. So we multiply both sides by 3\sqrt{3}, it will become log3x=23log_{3x} = 2\sqrt{3}. Then, we have (3x)23=10(3x)^{2\sqrt{3}} = 10. So, let's proceed with this equation.

Then, we raise 3x3x to the power of 232\sqrt{3}. This gives us: (3x)23=10(3x)^{2\sqrt{3}} = 10. This is where things get a bit tricky, but it's manageable. We need to solve for x. The equation is now (3x)23=10(3x)^{2\sqrt{3}} = 10. This equation is difficult to solve algebraically. To solve for x, we can make some approximations using numerical methods.

Let's first isolate 3x. For this, we'll take both sides to the power of 123\frac{1}{2\sqrt{3}}:

(3x)23βˆ—123=10123(3x)^{2\sqrt{3} * \frac{1}{2\sqrt{3}}} = 10^{\frac{1}{2\sqrt{3}}}

3x=101233x = 10^{\frac{1}{2\sqrt{3}}}.

Then, x=13βˆ—10123x = \frac{1}{3} * 10^{\frac{1}{2\sqrt{3}}}.

Using a calculator, 123β‰ˆ0.2887\frac{1}{2\sqrt{3}} \approx 0.2887. Then, 100.2887β‰ˆ1.94410^{0.2887} \approx 1.944. So,

xβ‰ˆ1.9443β‰ˆ0.648x \approx \frac{1.944}{3} \approx 0.648.

So, the approximate solution for x is about 0.648. Now, we've solved for x in the given logarithmic equation. It's really helpful to understand how each step works, isn't it? Practice will make perfect, and soon you'll be solving these with ease.

Important Note: Always check your solutions in the original equation to ensure they are valid. Remember that the argument of a logarithm (the part inside the log) must be positive, and the base must also be positive and not equal to 1.

Evaluating the Logarithm: c. log⁑333=?\log_{3} 3\sqrt{3} = ?

Alright, moving on to the second part of our task: evaluating the logarithm log⁑333\log_{3} 3\sqrt{3}. This is all about finding the exponent to which we must raise the base (3) to get the argument (333\sqrt{3}). Let's break this down step by step to make it super clear. Our goal is to find the value of the expression log⁑333\log_{3} 3\sqrt{3}.

First, let's rewrite the argument, which is 333\sqrt{3}. We can express this using exponents. Remember that 3=312\sqrt{3} = 3^{\frac{1}{2}}. Therefore, 33=31βˆ—3123\sqrt{3} = 3^1 * 3^{\frac{1}{2}}.

Now, when multiplying numbers with the same base, we add the exponents. So, 31βˆ—312=31+12=3323^1 * 3^{\frac{1}{2}} = 3^{1 + \frac{1}{2}} = 3^{\frac{3}{2}}.

So, we can rewrite the original expression as log⁑3332\log_{3} 3^{\frac{3}{2}}. The logarithm asks: "To what power must we raise 3 to get 3323^{\frac{3}{2}}?" The answer is clearly 32\frac{3}{2}.

Thus, log⁑333=32\log_{3} 3\sqrt{3} = \frac{3}{2}.

Alternatively, you can think of it like this: log⁑333=x\log_{3} 3\sqrt{3} = x. Then, 3x=333^x = 3\sqrt{3}.

3x=31βˆ—312=3323^x = 3^1 * 3^{\frac{1}{2}} = 3^{\frac{3}{2}}.

So, x=32x = \frac{3}{2}.

Therefore, the value of the logarithm log⁑333\log_{3} 3\sqrt{3} is 32\frac{3}{2} or 1.5. This kind of problem often appears in math tests, so understanding the steps is important. Remember, always try to simplify the argument of the logarithm into a power of the base, as it simplifies the process.

Key Concepts and Reminders

Let's quickly go over some key concepts and important reminders to help you master logarithmic equations and evaluations.

  • Logarithm Definition: log⁑ba=c\log_{b} a = c means that bc=ab^c = a. 'b' is the base, 'a' is the argument, and 'c' is the exponent (the value of the logarithm).
  • Exponential Form: Knowing how to switch between logarithmic form and exponential form is crucial for solving these problems. Always remember how the base, argument, and exponent relate to each other.
  • Properties of Logarithms: Remember the properties of exponents and logarithms, such as:
    • log⁑b(mn)=log⁑bm+log⁑bn\log_{b} (mn) = \log_{b} m + \log_{b} n
    • log⁑b(mn)=log⁑bmβˆ’log⁑bn\log_{b} (\frac{m}{n}) = \log_{b} m - \log_{b} n
    • log⁑bmn=nlog⁑bm\log_{b} m^n = n \log_{b} m
  • Base Restrictions: The base of a logarithm (b) must be positive and not equal to 1. The argument (a) must be positive.
  • Simplification: Always try to simplify the argument of the logarithm using exponent rules before evaluating.

Tips for Success

Here are a few tips to help you succeed in solving logarithmic equations and evaluating logarithms:

  • Practice Regularly: The more you practice, the more comfortable you'll become with the concepts. Work through a variety of examples.
  • Understand the Basics: Make sure you have a solid understanding of exponents and logarithms before tackling complex problems.
  • Use a Calculator Wisely: A scientific calculator can be very helpful, especially for simplifying expressions and evaluating logarithms. But make sure to understand the steps involved, not just rely on the calculator.
  • Check Your Answers: Always verify your solutions in the original equation to ensure they are valid. This helps catch any errors and reinforces your understanding.
  • Break It Down: Don't try to solve everything at once. Break complex problems into smaller, more manageable steps.

By following these tips and practicing consistently, you'll be well on your way to mastering logarithmic equations and evaluations. Keep up the great work, and don't be afraid to ask for help if you need it. Math can be fun, and with the right approach, you can definitely succeed!

I hope this step-by-step guide has been helpful, guys! Feel free to ask if anything is unclear, and keep practicing! You got this! Remember, it's all about understanding the concepts and building your skills gradually. Good luck with your math journey!