Problem 38 - Beginning and Intermediate Algebra

Question

Solve the logarithmic equation algebraically. Then check using a graphing calculator. $$\log _{125} \frac{1}{25}=x$$

Step-by-Step Solution

Answer

To solve the logarithmic equation \(\log _{125} \frac{1}{25}=x\), first convert it to exponential form: \(125^x = \frac{1}{25}\). Rewrite the bases as powers of 5: \((5^3)^x = 5^{-2}\). Simplify and solve for x: \(5^{3x} = 5^{-2}\), so \(3x = -2\) and \(x = \frac{-2}{3}\). Check the solution using a graphing calculator by graphing \(y = \log_{125} \frac{1}{25}\) and \(y = x\). The intersection point confirms the solution is \(x = -\frac{2}{3}\).

Step 1: Convert the logarithmic equation to exponential form

Recall that the logarithmic equation \(\log_b a = c\) is equivalent to the exponential equation \(b^c = a\). So, the given equation can be rewritten as: \[125^x = \frac{1}{25}\]

Step 2: Rewrite the bases as powers of a common base

Both 125 and 25 can be written as powers of 5. Specifically, \(125 = 5^3\) and \(25 = 5^2\). Substitute these expressions into the equation in step 1: \[(5^3)^x = 5^{-2}\]

Step 3: Simplify and solve for x

Using the exponent rule \((a^b)^c = a^{b*c}\), we simplify the left side of the equation: \[5^{3x} = 5^{-2}\] Since the bases are the same (both are 5), we can set the exponents equal to each other: \[3x = -2\] Now solve for x: \[x = \frac{-2}{3}\]

Step 4: Check the solution using a graphing calculator

To check the solution, graph \(y = \log_{125} \frac{1}{25}\) and \(y = x\) on a graphing calculator. The point of intersection between these two graphs will give the value of x. You should find that the point of intersection is \(\left(\frac{-2}{3}, \frac{-2}{3}\right)\), confirming that the solution \(x = -\frac{2}{3}\) is correct.

Key Concepts

Exponential Form

Converting a logarithmic equation to its exponential form can be a game-changer. This process simplifies the equation and reveals relationships we need to solve it. Remember that a logarithm is essentially the inverse of an exponentiation.
When you see a logarithmic equation such as \(\log_b a = c\), this implies that the base raised to the power of the logarithm equals the value. In equation terms, it’s rewritten as \(b^c = a\).
For instance, in our problem, the logarithm \(\log_{125} \frac{1}{25} = x\) becomes \(125^x = \frac{1}{25}\). This conversion allows us to see more readily how the values relate to each other through exponentiation.

Common Base

Finding a common base is key when dealing with exponentiation. It allows us to equate the exponents and simplifies the solving process.
Both numbers, 125 and 25 in the problem, can be expressed using a common base of 5:

• \(125 = 5^3\)
• \(25 = 5^2\)

Once we have these forms, we can rewrite the equation \(125^x = \frac{1}{25}\) as \((5^3)^x = 5^{-2}\). This representation with a common base simplifies the comparison between the left and right sides of the equation.

Exponent Rules

Exponent rules are vital to manipulate and solve equations efficiently. One essential rule used in our exercise is \((a^b)^c = a^{b \cdot c}\). This tells us how to handle exponents raised to another power.
Applying this to our problem, \((5^3)^x\) becomes \(5^{3x}\). By rewriting the equation \(5^{3x} = 5^{-2}\), and since the bases are identical, the exponents can be equated directly:

• \(3x = -2\)

Finally, solving this by isolating \(x\) gives \(x = \frac{-2}{3}\). Utilizing these rules makes exponent manipulation straightforward.