Solving the Exponential and Logarithmic Equation for x

We are given the equation \( 5^{x-1} = (2.5)^{\log_{10} 5} \) and our goal is to find the value of \( x \). This equation involves both exponential and logarithmic terms, so we will use properties of logarithms to solve it.

Step-by-Step Solution to Find x

To solve for \( x \) when the variable is in the exponent, taking the logarithm of both sides is a common technique. Using the base-10 logarithm (\(\log_{10}\)) is convenient here because of the \(\log_{10} 5\) term already present.

1. Start with the original equation: \[ 5^{x-1} = (2.5)^{\log_{10} 5} \]
2. Take the base-10 logarithm of both sides of the equation: \[ \log_{10}(5^{x-1}) = \log_{10}((2.5)^{\log_{10} 5}) \]
3. Apply the Power Rule of Logarithms, \( \log_b(M^p) = p \log_b M \), to both sides:

   The left side becomes \( (x-1) \log_{10} 5 \).

   The right side becomes \( (\log_{10} 5) \log_{10}(2.5) \).

   The equation is now:

   \[ (x-1) \log_{10} 5 = (\log_{10} 5) \log_{10}(2.5) \]
4. Isolate the term containing \( x \) by dividing both sides by \( \log_{10} 5 \):

   Since \( \log_{10} 5 \) is a non-zero value, we can safely divide by it.

   \[ x-1 = \log_{10}(2.5) \]
5. Simplify the logarithm term on the right side:

   Convert the decimal \( 2.5 \) into a fraction: \( 2.5 = \frac{5}{2} \).

   Apply the Quotient Rule of Logarithms, \( \log_b(M/N) = \log_b M - \log_b N \):

   \[ \log_{10}(2.5) = \log_{10}\left(\frac{5}{2}\right) = \log_{10} 5 - \log_{10} 2 \]

   Substituting this back into the equation for \( x-1 \):

   \[ x-1 = \log_{10} 5 - \log_{10} 2 \]
6. Solve for \( x \) by adding 1 to both sides: \[ x = 1 + \log_{10} 5 - \log_{10} 2 \]
7. Express the constant '1' as a base-10 logarithm:

   We know that \( \log_{10} 10 = 1 \).

   \[ x = \log_{10} 10 + \log_{10} 5 - \log_{10} 2 \]
8. Combine the logarithm terms using the Product and Quotient Rules:

   \( \log_{10} 10 + \log_{10} 5 = \log_{10}(10 \times 5) = \log_{10}(50) \)

   \[ x = \log_{10}(50) - \log_{10} 2 \]

   Now apply the Quotient Rule again:

   \[ x = \log_{10}\left(\frac{50}{2}\right) \] \[ x = \log_{10}(25) \]
9. Simplify the result further:

   Recognize that \( 25 \) can be written as \( 5^2 \).

   \[ x = \log_{10}(5^2) \]

   Apply the Power Rule of Logarithms one more time:

   \[ x = 2 \log_{10} 5 \]

The value of \( x \) that satisfies the given equation is \( 2 \log_{10} 5 \).

Revision Table: Key Logarithm Properties

Understanding these basic properties is fundamental for solving logarithmic and exponential equations.

• \( \log_b (MN) = \log_b M + \log_b N \) (Product Rule)
• \( \log_b (M/N) = \log_b M - \log_b N \) (Quotient Rule)
• \( \log_b (M^p) = p \log_b M \) (Power Rule)
• \( \log_b b = 1 \)
• \( \log_b 1 = 0 \)
• \( b^{\log_b M} = M \)

Additional Information: Solving Exponential Equations

Exponential equations are equations where the unknown variable appears as an exponent. Solving these often involves converting between exponential and logarithmic forms or taking logarithms of both sides.

• Method 1: Making Bases the Same

  If you can rewrite both sides of the equation with the same base, say \( a^f(x) = a^{g(x)} \), then you can simply set the exponents equal: \( f(x) = g(x) \). This method is straightforward but not always possible.

• Method 2: Using Logarithms (Taking Log of Both Sides)

  When the bases cannot be easily made the same, take the logarithm of both sides of the equation. You can use any convenient base (like base 10 or natural logarithm, ln). For \( a^{f(x)} = b^{g(x)} \):

  \( \log(a^{f(x)}) = \log(b^{g(x)}) \)

  \( f(x) \log a = g(x) \log b \) (Using Power Rule)

  Then, solve the resulting algebraic equation for \( x \). This is the method used in the solution above.