# Logarithm properties in functional equations Show that f(1/a) = -f(a)

• MHB
• My Name is Earl
In summary: Excellent! Now, for part (c), I would begin by using the definition of $$f$$ to write:f\left(\frac{a}{b}\right)=f\left(a\cdot\frac{1}{b}\right)=f(a)+f\left(\frac{1}{b}\right)Now, use the result from part (b)...what do you get?Yes good...and so you have derived $$P_{n+1}$$ from $$P_n$$, completing the proof by induction.
My Name is Earl
Suppose that we have a function f(x) such that f(ab) = f(a)+f(b) for all rational numbers a and b.
(a) Show that f(1) = 0.
(b) Show that f(1/a) = -f(a).
(c) Show that f(a/b) = f(a) - f(b).
(d) Show that f(an) = nf(a) for every positive integer a.

For (a), if ab = 1 then a = 1/b and b = 1/a. Not sure how to proceed from here.

Last edited:
My Name is Earl said:
Suppose that we have a function f(x) such that f(ab) = f(a)+f(b) for all rational numbers a and b.
(a) Show that f(1) = 0.
(b) Show that f(a/b) = -f(a).
(c) Show that f(a/b) = f(a) - f(b).
(d) Show that f(an) = nf(a) for every positive integer a.

For (a), if ab = 1 then a = 1/b and b = 1/a. Not sure how to proceed from here.

For part (a), suppose we let $$b=1$$...what does this imply from the given definition of $$f$$?

I think part (b) should be:

Show that:

$$\displaystyle f\left(\frac{1}{a}\right)=-f(a)$$

For this, using what you stated for part (a), that is to let $$ab=1$$, along with the result from part (a), will be a good strategy.

MarkFL said:
I think part (b) should be:

Show that:

$$\displaystyle f\left(\frac{1}{a}\right)=-f(a)$$

For this, using what you stated for part (a), that is to let $$ab=1$$, along with the result from part (a), will be a good strategy.

You are correct - fixed it in original post. Thank you.

MarkFL said:
For part (a), suppose we let $$b=1$$...what does this imply from the given definition of $$f$$?
If b = 1 then f(a) = f(a) + f(1) therefore f(1) = 0. Thanks

MarkFL said:
I think part (b) should be:

Show that:

$$\displaystyle f\left(\frac{1}{a}\right)=-f(a)$$

For this, using what you stated for part (a), that is to let $$ab=1$$, along with the result from part (a), will be a good strategy.

From part a: If ab = 1, then f(1) = 0 and b = 1/a
Substituting these values into f(ab) = f(a) + f(b)
0 = f(a) + f(1/a)
f(1/a) = -f(a)

My Name is Earl said:
From part a: If ab = 1, then f(1) = 0 and b = 1/a
Substituting these values into f(ab) = f(a) + f(b)
0 = f(a) + f(1/a)
f(1/a) = -f(a)

Excellent! Now, for part (c), I would begin by using the definition of $$f$$ to write:

$$\displaystyle f\left(\frac{a}{b}\right)=f\left(a\cdot\frac{1}{b}\right)=f(a)+f\left(\frac{1}{b}\right)$$

Now, use the result from part (b)...what do you get?

MarkFL said:
Excellent! Now, for part (c), I would begin by using the definition of $$f$$ to write:

$$\displaystyle f\left(\frac{a}{b}\right)=f\left(a\cdot\frac{1}{b}\right)=f(a)+f\left(\frac{1}{b}\right)$$

Now, use the result from part (b)...what do you get?

Let $$\displaystyle b = \frac{1}{b}$$, then substituting into f(ab) = f(a) + f(b):
$$\displaystyle f(a\cdot\frac{1}{b}) = f(a) + f(\frac{1}{b})$$
From part (b) it follows that $$\displaystyle f(\frac{1}{b}) = -f(b)$$, therefore
$$\displaystyle f(a\cdot\frac{1}{b}) = f(a) - f(b)$$

I would use an inductive prove for part (d). Consider the induction hypothesis $$P_n$$:

$$\displaystyle f\left(a^n\right)=nf(a)$$ where $$n\in\mathbb{N}$$

We see the base case $$P_1$$ is true, because it leads to the identity:

$$\displaystyle f(a)=f(a)$$

So, applying the given rule for the function and $$P_n$$, we can state:

$$\displaystyle f\left(a^{n+1}\right)=f\left(a\cdot a^{n}\right)=\,?$$

MarkFL said:
I would use an inductive prove for part (d). Consider the induction hypothesis $$P_n$$:

$$\displaystyle f\left(a^n\right)=nf(a)$$ where $$n\in\mathbb{N}$$

We see the base case $$P_1$$ is true, because it leads to the identity:

$$\displaystyle f(a)=f(a)$$

So, applying the given rule for the function and $$P_n$$, we can state:

$$\displaystyle f\left(a^{n+1}\right)=f\left(a\cdot a^{n}\right)=\,?$$

continuing from above...

$$\displaystyle = f(a) + f(a^{n})$$
$$\displaystyle = f(a) + n\cdot f(a)$$
$$\displaystyle = (n+1) \cdot f(a)$$

My Name is Earl said:
continuing from above...

$$\displaystyle = f(a) + f(a^{n})$$
$$\displaystyle = f(a) + n\cdot f(a)$$
$$\displaystyle = (n+1) \cdot f(a)$$

Yes good...and so you have derived $$P_{n+1}$$ from $$P_n$$, completing the proof by induction. (Yes)

## 1. What is a logarithm?

A logarithm is a mathematical function that represents the exponent or power to which a base number must be raised to produce a given number.

## 2. What are the properties of logarithms?

There are several properties of logarithms, including the product rule, quotient rule, power rule, and change of base rule. These properties allow for the simplification and manipulation of logarithmic expressions.

## 3. How do logarithms relate to functional equations?

In functional equations, logarithmic expressions can be used to represent the relationship between the independent variable (input) and the dependent variable (output) in a function. Logarithms can also be used to solve for unknown variables in functional equations.

## 4. How can logarithm properties be used to show that f(1/a) = -f(a) in a functional equation?

By applying the quotient rule for logarithms, we can transform the functional equation f(1/a) = f(a) into -f(a) = f(a). Then, by subtracting f(a) from both sides, we get -2f(a) = 0. Finally, dividing both sides by -2 yields f(a) = 0, which proves the original equation f(1/a) = -f(a).

## 5. Can logarithm properties be used to solve other types of functional equations?

Yes, logarithm properties can be applied to various types of functional equations, including linear, quadratic, and exponential functions. By manipulating and simplifying the equations using logarithm properties, we can solve for unknown variables and find solutions to the functional equations.

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