Math Problem Statement

Solve for x in the equation x = 5^{\log_5(7) + \log_{\frac{1}{5}}(3)}.

Solution

The equation given is:

x=5log5(7)+log15(3)x = 5^{\log_5(7) + \log_{\frac{1}{5}}(3)}

Let's simplify this step by step.

  1. Simplify the first logarithm:
    log5(7)\log_5(7) remains as it is.

  2. Simplify the second logarithm:
    log15(3)\log_{\frac{1}{5}}(3) can be rewritten using the change of base formula: log15(3)=log5(3)log5(15)\log_{\frac{1}{5}}(3) = \frac{\log_5(3)}{\log_5\left(\frac{1}{5}\right)} We know that log5(15)\log_5\left(\frac{1}{5}\right) simplifies to 1-1 (since 51=155^{-1} = \frac{1}{5}). Therefore: log15(3)=log5(3)\log_{\frac{1}{5}}(3) = -\log_5(3)

  3. Combine the logarithms: Now the expression becomes: x=5log5(7)log5(3)x = 5^{\log_5(7) - \log_5(3)}

  4. Use the logarithmic property:
    The expression log5(7)log5(3)\log_5(7) - \log_5(3) can be simplified using the logarithm subtraction rule: log5(73)\log_5\left(\frac{7}{3}\right)

    So, the expression simplifies to: x=5log5(73)x = 5^{\log_5\left(\frac{7}{3}\right)}

  5. Final simplification:
    Since 5log5(73)5^{\log_5\left(\frac{7}{3}\right)} is simply 73\frac{7}{3}, we have: x=73x = \frac{7}{3}

So, the solution is:

73\boxed{\frac{7}{3}}

Would you like more details or have any other questions?


Here are 5 related questions:

  1. How do you simplify logarithms with different bases?
  2. What are the properties of logarithms that allow you to combine them?
  3. How do you convert between logarithms of different bases?
  4. What happens when the base of the logarithm is a fraction?
  5. How does the change of base formula work in logarithms?

Tip:

Remember that logarithms are powerful tools for simplifying complex expressions, especially when they involve exponents. Understanding the fundamental properties of logarithms will help you manipulate and solve such equations effectively.

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Math Problem Analysis

Mathematical Concepts

Logarithms
Exponents
Algebra

Formulas

Change of Base Formula: \log_b(a) = \frac{\log_c(a)}{\log_c(b)}
Logarithm Subtraction Rule: \log_b(a) - \log_b(c) = \log_b\left(\frac{a}{c}\right)

Theorems

Properties of Logarithms
Change of Base Theorem

Suitable Grade Level

Grades 9-12