Question

Prove that $({\log }_{b}c)({\log }_{c}b)=1,$ for $b>0$ $c>0,b\ne 1,$ and $c\ne 1$

Short answer

Question: Prove that $({\log }_{b}c)({\log }_{c}b)=1$. Answer: We can prove this by applying the change-of-base formula. We rewrite the expression as $\frac{{\log }_{c}c}{{\log }_{c}b}\cdot \frac{{\log }_{c}b}{{\log }_{c}c}$, which simplifies to $\frac{1}{{\log }_{c}b}\cdot \frac{{\log }_{c}b}{1}$. The common factors cancel out, leaving us with 1. Therefore, $({\log }_{b}c)({\log }_{c}b)=1$.

Step-by-step solution

Write down the given expression

We are given the expression: $({\log }_{b}c)({\log }_{c}b)$

Apply change-of-base formula

The change-of-base formula states that ${\log }_{a}b=\frac{{\log }_{c}b}{{\log }_{c}a}$ for any positive choice of $c$ not equal to 1. We can apply this formula in our expression with $c$ as the new base, which results in: $({\log }_{b}c)({\log }_{c}b)=\frac{{\log }_{c}c}{{\log }_{c}b}\cdot \frac{{\log }_{c}b}{{\log }_{c}c}$

Simplify the expression

We know that ${\log }_{a}a=1$ for any base $a$. So, we can replace ${\log }_{c}c$ with 1: $\frac{1}{{\log }_{c}b}\cdot \frac{{\log }_{c}b}{1}$ Now, we can cancel out the common factors in the numerator and denominator: $\frac{\cancel{{\log }_{c}b}}{\cancel{{\log }_{c}b}}=1$

Conclusion

We have now proved that $({\log }_{b}c)({\log }_{c}b)=1$ for any positive choices of $b$ and $c$ not equal to 1.

Key concepts

Change-of-Base Formula

The change-of-base formula is a powerful tool in mathematics, particularly in logarithms. It allows us to compute a logarithm with a different base that might be more convenient for calculations. The formula is given by

• ${\log }_{a}b=\frac{{\log }_{c}b}{{\log }_{c}a}$

Here, $a$, $b$, and $c$ are all greater than 0, and neither $a$ nor $c$ can be 1.
This formula is useful for several reasons, including simplifying complex logarithmic expressions.
In the context of the problem, the change-of-base formula is applied to break down

• ${\log }_{b}c$
• ${\log }_{c}b$

into a form that allows straightforward simplification. By using a common base, the calculations further become more accessible.
The change-of-base formula conveniently transforms logarithmic expressions to any base that suits the problem, enhancing our ability to manipulate and solve them.

Simplification in Algebra

Simplification in algebra often involves reducing expressions to their simplest form where they are easily understandable and solvable.
In the context of logarithms, simplification may include cancelling out terms, applying identities, and using properties to combine or reduce expressions.
For instance, in simplifying the expression

• $\frac{{\log }_{c}c}{{\log }_{c}b}\cdot \frac{{\log }_{c}b}{{\log }_{c}c}$

we can see a common factor in both the numerator and the denominator.
The property of logarithms tells us that ${\log }_{a}a=1$, which was crucial in simplifying our expression, highlighting the elegance of simplification.
This process boils down complicated expressions using mathematics's inherent rules, making them much more feasible to handle.

Properties of Logarithms

Logarithms have several key properties that make them very useful in algebraic manipulations.
Some important properties include the product, quotient, power, and change-of-base properties.
These facilitate simplification of logarithmic expressions, especially during complex calculations.

• The product property states that ${\log }_b(xy)={\log }_{b}x+{\log }_{b}y$.
• The quotient property is ${\log }_b\left(\frac{x}{y}\right)={\log }_{b}x-{\log }_{b}y$.
• The power property, ${\log }_b(x^c)=c\cdot {\log }_{b}x$, helps in handling exponential terms.

In application, knowing that ${\log }_{a}a=1$ allows for simplification. It was directly used in the exercise to reduce the expression to 1.
Using these properties enables the breakdown of complex expressions into simpler components, making calculations much more efficient.