Problem 59 - Intermediate Algebra

Question

Solve for \(x .\) See Example 3. $$ \log _{36} x=-\frac{1}{2} $$

Step-by-Step Solution

Answer

\( x = \frac{1}{6} \).

Step 1: Understanding Logarithmic Equations

The equation given is \( \log_{36} x = -\frac{1}{2} \). This is a logarithmic equation, where \( \log_{b} a = c \) means \( b^{c} = a \). In this case, \( b = 36 \), \( c = -\frac{1}{2} \), and \( a = x \). This means we need to convert the logarithmic equation into an exponential form.

Step 2: Converting Logarithmic to Exponential Form

From \( \log_{36} x = -\frac{1}{2} \), we convert it to exponential form using the definition of logarithms: \( 36^{-\frac{1}{2}} = x \). This means \( 36^{-\frac{1}{2}} \) is equal to \( x \).

Step 3: Simplify the Exponential Expression

To compute \( 36^{-\frac{1}{2}} \), we recognize that a negative exponent represents the reciprocal, and the \(-\frac{1}{2}\) indicates a square root. So, \( 36^{-\frac{1}{2}} = \frac{1}{\sqrt{36}} \).

Step 4: Calculate the Square Root

The next step is to find \( \sqrt{36} \). We know that \( 36 \) is a perfect square, and \( \sqrt{36} = 6 \).

Step 5: Final Calculation

Substituting back, we have \( 36^{-\frac{1}{2}} = \frac{1}{6} \). This means \( x = \frac{1}{6} \).

Key Concepts

Understanding Exponential Form

When dealing with logarithmic equations, converting them into exponential form is a crucial step. Logarithms are the inverse operations of exponentiation.
The basic idea is that if you have a logarithm equation like \( \log_b a = c \), it can be rewritten in its exponential form as \( b^c = a \). This transformation helps in visualizing the equation in terms of exponents, which can often be easier to manipulate and solve.
In the given example, \( \log_{36} x = -\frac{1}{2} \) transforms to \( 36^{-\frac{1}{2}} = x \).
This step is key as it moves us from the logarithmic understanding to working directly with powers and exponents.

Exploring Negative Exponents

Negative exponents can be initially confusing, but they are quite straightforward once you understand the concept. A negative exponent in an expression indicates the reciprocal of the base raised to the absolute value of the exponent.
For example, \( b^{-n} = \frac{1}{b^n} \). In our exercise, we encounter \( 36^{-\frac{1}{2}} \), which is interpreted as \( \frac{1}{36^{\frac{1}{2}}} \).
Writing it this way allows us to apply the idea that the negative exponent reverses the traditional exponent effect into a division by that exponential value.

Understanding Square Roots

Square roots are another essential concept when dealing with exponents and logarithms.
They allow us to solve equations where exponents involve fractions, such as \( \frac{1}{2} \).
The square root symbol \( \sqrt{} \) is applied to find the number which, when multiplied by itself, will produce the original number.
In our specific case of \( 36^{\frac{1}{2}} \), we know this to be \( \sqrt{36} \), which equals 6.
So the expression \( \frac{1}{\sqrt{36}} \) equates to \( \frac{1}{6} \), hence solving for the value of \( x \), resulting in \( x = \frac{1}{6} \).
This demonstrates how logarithms, exponents, and roots combine to simplify seemingly complicated equations.