Problem 73 - Algebra 2 and Trigonometry

Question

Solve for \(a : \log _{5} 0.2=\log _{a} 10\)

Step-by-Step Solution

Answer

The value of \(a\) is 0.1.

Step 1: Understanding the Given Equation

We have the equation \( \log_{5} 0.2 = \log_{a} 10 \). This equation shows that the logarithms with different bases and different arguments are equal, which implies that the bases are correctly inverted powers of each other for equality.

Step 2: Expressing via Logarithmic Identity

Use the change of base formula, \( \log_{b}(c) = \frac{\log(c)}{\log(b)} \), on both sides of the equation to express them in terms of common logarithms (base 10).We have:\[ \frac{\log(0.2)}{\log(5)} = \frac{\log(10)}{\log(a)} \]

Step 3: Simplifying Using Logarithmic Properties

We know \( \log(10) = 1 \) since 10 is the base of common logarithms. Use the identity \( \log(0.2) = \log(\frac{1}{5}) \) which can be rewritten using the property of logarithms as \( \log(1) - \log(5) \) or \( 0 - \log(5) = -\log(5) \). Substitute these into the equation:\[ \frac{-\log(5)}{\log(5)} = \frac{1}{\log(a)} \]

Step 4: Solving the Simplified Equation

Simplify \( \frac{-\log(5)}{\log(5)} = -1 \) and set it equal to \( \frac{1}{\log(a)} \).Thus, the equation becomes:\[ -1 = \frac{1}{\log(a)} \]

Step 5: Finding \(\log(a)\)

To solve \(-1 = \frac{1}{\log(a)}\), rearrange to find \(\log(a) = -1\).

Step 6: Calculating the Value of \(a\)

Recall that \( \log(a) = -1 \) means that \( a \) is equal to the base raised to the power of \(-1\), i.e., \( a = 10^{-1} \), which simplifies to \( a = 0.1 \).

Key Concepts

Change of Base Formula

The change of base formula is a handy tool in logarithms, particularly when dealing with different bases, as in the equation given in the original exercise. It helps us convert logarithms with any base to a logarithm with a base that may be easier to work with, like base 10 (common logarithms) or base e (natural logarithms). This is essential because calculators and most computational tools readily provide values for logarithms of these bases.

Using the change of base formula, you can express \(\log_{b}(c)\) as:

• \( \log_{b}(c) = \frac{\log(c)}{\log(b)} \)

This formula means that any logarithm can be transformed to a ratio of common logarithms. In the exercise, we used this formula to transform both sides of the equation \(\log_5(0.2) \) and \(\log_a(10)\) into a form that uses base 10 logarithms.

This transformation enables simplification and easier manipulation when solving the equation, as base 10 logarithms are often pre-tabulated or can be approximated with known values.

Logarithmic Identities

Logarithmic identities are fundamental in simplifying expressions and solving logarithmic equations. These identities are rooted in the definition and properties of logarithms. In the given exercise, we used a couple of these identities to simplify expressions.

One such identity is understanding that:

• \( \log(1) = 0 \)

This comes from the fact that any number to the power of zero is 1. Another important identity, used in the exercise, is the reciprocal property:\( \log(c^{-1}) = -\log(c) \). This means when you have a logarithm of a reciprocal, you can simplify it by changing the sign of the logarithm.

For instance, we knew that \(0.2 = \frac{1}{5}\), which allowed us to express \(\log(0.2)\) as \(-\log(5)\). These kinds of transformations are key in manipulating and solving logarithmic equations, as they align both sides to a format that is easily comparable.

Logarithmic Properties

The properties of logarithms are tools that enable deeper insights and manipulations when dealing with logarithmic equations. In our solution, we applied a specific property known as the power property of logarithms:

• \( b^{\log_b(x)} = x \)

This property, while not explicitly used in the original exercise, is fundamentally grounded in the solving process where the inverse nature of exponentials and logarithms comes into play.

We also leaned on the property that allows conversion between logarithmic and exponential forms, where if \( \log(a) = b \), then \(a = 10^b\) when dealing with base 10 logarithms. This was pivotal in concluding \(a = 10^{-1}\) from the previously found \( \log(a) = -1 \).

The logarithmic properties enable important steps such as rearranging terms and simplifying complex expressions, making the solving of logarithmic equations more straightforward and intuitive.