Solve Log ⁡ 2 X = Log ⁡ 5 3 + 1 \log_2 X = \log_5 3 + 1 Lo G 2 ​ X = Lo G 5 ​ 3 + 1 By Graphing. Round To The Nearest Tenth.$x = [ \square $

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Introduction

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In this article, we will explore the concept of logarithms and how to solve equations involving logarithms using graphing. We will start by understanding the properties of logarithms and then move on to solving the given equation.

Properties of Logarithms

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A logarithm is the inverse operation of exponentiation. It is a mathematical function that takes a number as input and returns the power to which a base number must be raised to produce that number. For example, $\log_2 8 = 3$ because $2^3 = 8$. The base of a logarithm is the number that is being raised to a power, and the argument of a logarithm is the number that is being taken as the power.

There are several properties of logarithms that are useful to know when solving equations involving logarithms. These properties include:

• Product Rule: $\log_b (xy) = \log_b x + \log_b y$
• Quotient Rule: $\log_b \left(\frac{x}{y}\right) = \log_b x - \log_b y$
• Power Rule: $\log_b x^y = y \log_b x$

Solving the Equation

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The given equation is $\log_2 x = \log_5 3 + 1$. To solve this equation, we can start by using the properties of logarithms to simplify the right-hand side.

$\log_2 x = \log_5 3 + 1$

Using the Power Rule, we can rewrite the right-hand side as:

$\log_2 x = \log_5 3 + \log_2 2$

Since $\log_2 2 = 1$, we can simplify the right-hand side further:

$\log_2 x = \log_5 3 + 1$

Now, we can use the Product Rule to rewrite the right-hand side as:

$\log_2 x = \log_5 3 \cdot \log_2 2$

Since $\log_2 2 = 1$, we can simplify the right-hand side further:

$\log_2 x = \log_5 3$

Graphing the Functions

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To solve the equation, we can graph the functions $y = \log_2 x$ and $y = \log_5 3$ on the same coordinate plane. The point of intersection of the two graphs will give us the value of $x$ that satisfies the equation.

Graphing $y = \log_2 x$

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The graph of $y = \log_2 x$ is a logarithmic curve that opens upwards. The curve passes through the point $(1, 0)$ and has a vertical asymptote at $x = 0$.

Graphing $y = \log_5 3$

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The graph of $y = \log_5 3$ is a horizontal line that passes through the point $(1, \log_5 3)$.

Finding the Point of Intersection

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To find the point of intersection of the two graphs, we can set the two functions equal to each other and solve for $x$.

$\log_2 x = \log_5 3$

Using the Change of Base Formula, we can rewrite the equation as:

$\frac{\log x}{\log 2} = \frac{\log 3}{\log 5}$

Cross-multiplying, we get:

$\log x \log 5 = \log 3 \log 2$

Dividing both sides by $\log 5$, we get:

$\log x = \frac{\log 3 \log 2}{\log 5}$

Using the Power Rule, we can rewrite the right-hand side as:

$\log x = \log 3^{\frac{\log 2}{\log 5}}$

Since $\log x = \log 3^{\frac{\log 2}{\log 5}}$, we can conclude that:

$x = 3^{\frac{\log 2}{\log 5}}$

Evaluating the Expression

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To evaluate the expression $x = 3^{\frac{\log 2}{\log 5}}$, we can use a calculator to find the value of the exponent.

$x \approx 3^{0.43067}$

Using the Power Rule, we can rewrite the expression as:

$x \approx 3^{0.43067} \approx 3 \cdot 3^{0.43067 - 1}$

$x \approx 3 \cdot 3^{-0.56933}$

$x \approx 3 \cdot \frac{1}{3^{0.56933}}$

$x \approx 3 \cdot \frac{1}{3^{0.56933}}$

$x \approx 3 \cdot 0.55555$

$x \approx 1.66667$

Rounding to the nearest tenth, we get:

$x \approx 1.7$

The final answer is: $\boxed{1.7}$

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Frequently Asked Questions

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Q: What is the main concept of this article?

A: The main concept of this article is to solve the equation $\log_2 x = \log_5 3 + 1$ using graphing.

Q: What are the properties of logarithms that are used in this article?

A: The properties of logarithms used in this article include the Product Rule, Quotient Rule, and Power Rule.

Q: How do you graph the functions $y = \log_2 x$ and $y = \log_5 3$?

A: The graph of $y = \log_2 x$ is a logarithmic curve that opens upwards, passing through the point $(1, 0)$ and having a vertical asymptote at $x = 0$. The graph of $y = \log_5 3$ is a horizontal line that passes through the point $(1, \log_5 3)$.

Q: How do you find the point of intersection of the two graphs?

A: To find the point of intersection of the two graphs, you can set the two functions equal to each other and solve for $x$.

Q: What is the value of $x$ that satisfies the equation?

A: The value of $x$ that satisfies the equation is $x = 3^{\frac{\log 2}{\log 5}}$.

Q: How do you evaluate the expression $x = 3^{\frac{\log 2}{\log 5}}$?

A: To evaluate the expression $x = 3^{\frac{\log 2}{\log 5}}$, you can use a calculator to find the value of the exponent.

Q: What is the final answer to the equation?

A: The final answer to the equation is $x \approx 1.7$.

Additional Questions and Answers

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Q: What is the significance of the Change of Base Formula in this article?

A: The Change of Base Formula is used to rewrite the equation $\log_2 x = \log_5 3$ in terms of common logarithms.

Q: How do you use the Power Rule to simplify the expression $x = 3^{\frac{\log 2}{\log 5}}$?

A: You can use the Power Rule to rewrite the expression as $x = 3^{\frac{\log 2}{\log 5}} = 3^{\log 2^{\frac{1}{\log 5}}} = 2^{\frac{1}{\log 5}}$.

Q: What is the relationship between the graphs of $y = \log_2 x$ and $y = \log_5 3$?

A: The graph of $y = \log_5 3$ is a horizontal line that passes through the point $(1, \log_5 3)$, while the graph of $y = \log_2 x$ is a logarithmic curve that opens upwards, passing through the point $(1, 0)$ and having a vertical asymptote at $x = 0$.

Q: How do you use the Product Rule to simplify the expression $\log_2 x = \log_5 3 + 1$?

A: You can use the Product Rule to rewrite the expression as $\log_2 x = \log_5 3 \cdot \log_2 2$.

Q: What is the significance of the Quotient Rule in this article?

A: The Quotient Rule is used to rewrite the expression $\log_2 x = \log_5 3 + 1$ in terms of common logarithms.

Conclusion

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In this article, we have solved the equation $\log_2 x = \log_5 3 + 1$ using graphing. We have used the properties of logarithms, including the Product Rule, Quotient Rule, and Power Rule, to simplify the expression and find the value of $x$ that satisfies the equation. We have also used the Change of Base Formula to rewrite the equation in terms of common logarithms. The final answer to the equation is $x \approx 1.7$.