Question

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

Short answer

Question: Prove that for any positive value of b and b ≠ 1, the following equation holds: \(\log _{1 / b} x=-\log _{b} x\). Answer: We proved that the given equation \(\log _{1 / b} x=-\log _{b} x\) holds true for all positive values of \(b\) and \(b \neq 1\) using the properties of logarithms and the change-of-base formula.

Step-by-step solution

Rewrite the given equation using the base \(a\)

Rewrite the given equation \(\log _{1 / b} x=-\log _{b} x\) using a common base, say base \(a\). Using the change-of-base formula, we get: $$\frac{\log _{a} x}{\log _{a}\left(1/b\right)}=-\frac{\log _{a} x}{\log _{a} b}$$

Simplify the equation using properties of logarithms

Using the properties of logarithms (\(\log{a^n}=n \log a\)), we simplify the equation further: $$\frac{\log _{a} x}{\log _{a} b^{-1}}=-\frac{\log _{a} x}{\log _{a} b}$$ Applying the property \(\log_a{b^{-1}}=-\log_a{b}\), we can simplify the equation as: $$\frac{\log _{a} x}{-\log _{a} b}=-\frac{\log _{a} x}{\log _{a} b}$$

Equate and cancel out the terms

Now we will equate the terms and cancel out: $$-\left(\frac{\log _{a} x}{\log _{a} b}\right)=-\frac{\log _{a} x}{\log _{a} b}$$ Now, we cancel \(-\log_a{x}\) from both sides and get: $$\frac{\log _{a} x}{\log _{a} b^{-1}}=\frac{\log _{a} x}{\log _{a} b}$$

Conclude the proof

In conclusion, the given equation \(\log _{1 / b} x=-\log _{b} x\) holds true for all positive values of \(b\) and \(b \neq 1\), as shown above.

Key concepts

Change-of-Base Formula

The change-of-base formula is an incredibly useful tool in solving logarithmic equations, particularly when you encounter a logarithm base that is not convenient to work with. Imagine you are performing calculations with a calculator that only computes logarithms to the base 10 or the natural logarithm base, e. In such cases, the change-of-base formula comes to the rescue, allowing you to convert the logarithm of any base to those more common bases.

Here's how the change-of-base formula looks: \[ \log_b(x) = \frac{\log_a(x)}{\log_a(b)} \].It asserts that the logarithm of a number x with base b can be expressed as the ratio of two logarithms with a new base a. The beauty of this formula lies in its versatility; a can be any positive value, not equal to 1, and still produce the correct result. For example, converting logarithms to base 10 would involve setting a to 10 in the formula. This property comes in especially handy when you're dealing with logarithmic equations where the bases are fractions or irrational numbers.

Properties of Logarithms

Logarithms are powerful mathematical tools, but their true potential is unlocked when you manipulate them using various properties. These properties are not just rules; they're insights into the nature of logarithms that simplify complex problems. Among the fundamental properties are:

• The Product Property: \( \log_b(xy) = \log_b(x) + \log_b(y) \), which allows you to break down the log of a product into a sum of logs.
• The Quotient Property: \( \log_b(\frac{x}{y}) = \log_b(x) - \log_b(y) \), useful for turning the logarithm of a quotient into a difference.
• The Power Property: \( \log_b(x^n) = n \cdot \log_b(x) \), empowering you to move the exponent of the argument in front as a multiplier.

These properties are not just arithmetic conveniences; they reflect the deeper logarithmic relationships that support solving equations algebraically. They also facilitate the transformation of equations into forms that are more tractable, a vital strategy when solving logarithmic equations.

Logarithmic Equations

Logarithmic equations involve unknowns within the log function, and solving them requires more than simple arithmetic—it’s a careful dance of applying properties of logarithms and understanding the underlying principles.

For example, consider the equation \( \log_b(x) = \log_b(y) \). If the bases are equal and the logs do not violate any logarithmic principles (like having a negative argument), one can deduce that the arguments should also be equal, hence \( x=y \).

But when the equations get trickier, with multiple logarithmic terms or varying bases, that's when you employ properties of logarithms alongside the change-of-base formula to simplify and solve the equations systematically. Solving logarithmic equations often involves isolating the logarithmic term, transforming the equation using properties, and then perhaps raising both sides as exponents to 'remove' the logarithms, allowing you to find the value of the unknown. With practice, patience, and precision, the complexities of logarithmic equations unfold into solvable algebraic expressions.