Problem 37 - Beginning and Intermediate Algebra

Question

Solve the logarithmic equation algebraically. Then check using a graphing calculator. $$\log _{64} \frac{1}{4}=x$$

Step-by-Step Solution

Answer

The solution for the logarithmic equation \(\log _{64} \frac{1}{4}=x\) is \(x=-\frac{1}{3}\). This can be verified by graphing the function on a calculator and checking that the x-intercept is equal to -1/3.

Step 1: Rewrite the equation in exponent form

We have the equation \(\log _{64} \frac{1}{4}=x\). To solve for x, we will rewrite the equation in exponent form using the definition of logarithm: \[\text{If} \; \log_{b}{a} = c, \; \text{then} \; b^c = a\] Rewriting the given equation in exponent form, we get: \[64^x = \frac{1}{4}\]

Step 2: Express both sides as powers of a common base

Now our equation is in the form \(64^x = \frac{1}{4}\). We can express both sides of the equation as powers with a common base to simplify. We can express 64 as \(4^3\) and \(\frac{1}{4}\) as \(4^{-1}\). The equation becomes: \[(4^3)^x = 4^{-1}\]

Step 3: Simplify and solve for x

Using the power of a power rule \((a^m)^n = a^{mn}\), we can rewrite and simplify the equation as: \[4^{3x} = 4^{-1}\] Since the bases are equal (4), we can set the exponents equal to each other: \[3x = -1\] Now solve for x: \[\frac{-1}{3} = x\] So, the solution for the equation \(\log _{64} \frac{1}{4}=x\) is: \[x = -\frac{1}{3}\]

Step 4: Check the solution using a graphing calculator

To check our solution, we can graph the function \(y = \log _{64} \frac{1}{4}\) using a graphing calculator. The x-intercept should be equal to our algebraic solution, which is -1/3. If the x-intercept is -1/3, then our solution is correct.

Key Concepts

Exponent Form

Understanding exponent form is crucial when dealing with logarithmic equations. In simple terms, a logarithm answers the question: "To what exponent must the base be raised, to produce a given number?" For instance, if you have \(\log_{b}{a} = c\), this means the base \( b \) raised to the power \( c \) equals \( a \). In our exercise, we start with the logarithmic equation \( \log_{64} \frac{1}{4}=x \), and convert it to exponent form to simplify. This becomes \( 64^x = \frac{1}{4} \). By rewriting in exponent form, we make it easier to solve for unknowns using algebraic methods.

Common Base

The idea of a common base helps simplify equations with exponents. In our equation \( 64^x = \frac{1}{4}\), different bases make solving complex. But, by converting both expressions to the same base, solutions become straightforward. Here, 64 is expressed as \( 4^3 \) and \( \frac{1}{4} \) as \( 4^{-1} \). Thus, the equation transforms into \((4^3)^x = 4^{-1}\). When both sides share the same base, you can equate the exponents directly, simplifying to solve for \( x \). Using a common base streamlines comparisons, making calculations much simpler and solving faster.

Power of a Power Rule

The power of a power rule is a key principle when dealing with exponents. It states that when you raise a power to another power, you multiply the exponents: \( (a^m)^n = a^{mn} \). This rule is applied in our equation \((4^3)^x = 4^{-1}\). By using the power of a power rule, we simplify it to \( 4^{3x} = 4^{-1} \). With the bases being same, we equate \( 3x = -1 \), allowing us to solve for \( x \): \( x = -\frac{1}{3} \). The power of a power rule is therefore instrumental in breaking down more complex exponential expressions to manageable steps.

Graphing Calculator

Graphing calculators are powerful tools to verify solutions visually. When you have solved an equation like \( \log _{64} \frac{1}{4}=x \) algebraically, you can double-check this by graphing. Input \( y = \log_{64} \left(\frac{1}{4}\right)\), and look for the x-intercept. The x-intercept displays where the graph crosses the x-axis, indicating the solution for x, which should match your algebraic result of \( x = -\frac{1}{3} \). By visualizing solutions, graphing calculators offer a clear way to confirm accuracy and provide further insight into the behavior of functions and relations between variables.