Question

Assume that \(b > 0\) and \(b \neq 1\). Show that \(\log _{1 / b} x=-\log _{b} x\).

Short answer

Question: Prove the logarithmic identity \(\log_{1/b} x = -\log_b x\), given that \(b > 0\) and \(b \neq 1\). Answer: The given logarithmic identity, \(\log_{1/b} x = -\log_b x\), has been proved using properties of logarithms and exponentials. We found that both sides of the identity have the same exponential equation, \(b^{RHS} = x\), indicating the equality of their exponents. Thus, the logarithmic identity holds true.

Step-by-step solution

Write the given identity as an equation

The given identity is \(\log_{1/b} x = -\log_b x\). Let us denote the left hand side as \(LHS\) and the right hand side as \(RHS\), so we have \(LHS = RHS\).

Convert the logarithms to exponentials

Recall that if \(\log_a y = x\), then \(a^x = y\). Using this property, we will convert the logarithmic equation to exponentials. For the \(LHS\), we have: \((1/b)^{LHS} = x\). For the \(RHS\), we have: \(b^{RHS} = x\).

Use the given identity in the exponential equation

We know that \(LHS = RHS\), so we can substitute \(RHS\) for \(LHS\) in the exponential equation for the LHS: \((1/b)^{RHS} = x\).

Use the reciprocal property in the exponential equation

Since \(1/b\) is the reciprocal of \(b\), we can write \((1/b)^{RHS}\) as \(b^{-RHS}\): \(b^{-RHS} = x\).

Compare both exponential equations

Now we have two exponential equations we found in step 2: 1. \(b^{RHS} = x\). 2. \(b^{-RHS} = x\). Since both equations have the same right side, it means that their left sides are equal as well: \(b^{RHS} = b^{-RHS}\).

Equate exponents

Since the bases in the equation \(b^{RHS} = b^{-RHS}\) are the same, it means their exponents must be equal: \(RHS = -RHS\).

Conclusion

Now that we proved that \(RHS = -RHS\), we can write the original identity as: \(\log_{1/b} x = -\log_b x\). The given logarithmic identity has been proved.

Key concepts

Exponential Functions

Exponential functions are mathematical expressions where a fixed base is raised to a variable exponent. They are crucial in understanding how quantities grow or decay rapidly. The general form of an exponential function is \(b^x\), where \(b\) is the base and \(x\) is the exponent.
In our problem, we encounter exponential functions when converting logarithms to their exponential form. For example, converting \(\log_a y = x\) to an exponential function results in \(a^x = y\). This transformation helps us solve equations involving logarithms by interpreting them as equations involving exponents.
Exponential functions are powerful because they describe many real-world phenomena, like population growth and radioactive decay. Understanding them is key to deciphering complex logarithmic identities and equations.

Reciprocal Property

The reciprocal property is essential when working with exponential and logarithmic equations. It tells us how to deal with numbers that are inverses of each other. Specifically, if \(b\) is a number, then its reciprocal is \(1/b\).
In the context of exponentiation, using the reciprocal property allows us to express \((1/b)^x\) as \(b^{-x}\). This transformation is particularly useful in simplifying complex expressions and solving equations, like in our given problem, where it helps transition from one form of an equation to another.

• Remember that the reciprocal property is not just a trick—it’s fundamental for simplifying mathematical expressions.
• It enables the conversion of a division into a multiplication, which can simplify many computation steps.

By understanding and applying the reciprocal property, we can more easily manipulate and solve mathematical identities involving both exponential and logarithmic functions.

Logarithmic Functions

Logarithmic functions are inverses of exponential functions. They help solve equations where the variable appears as an exponent. The logarithm \(\log_b x\) asks the question: "To what power must \(b\) be raised, to produce \(x\)?" This function is written as \(x = b^y\), which converts to \(y = \log_b x\).
Logarithmic identities, like the one in our exercise, often rely on understanding these functions deeply. For instance, \(\log_{1/b} x = -\log_b x\) takes advantage of reciprocal properties to show relationships between different bases.
Here's what makes logarithmic functions useful:

• They simplify multiplication into addition, division into subtraction—properties that are very powerful for calculations.
• They are fundamental in scientific calculations, like determining pH in chemistry or measuring sound intensity in decibels.

Grasping how to manipulate and use logarithmic functions allows for solving a broad range of mathematical problems efficiently, especially when paired with their exponential counterparts.