Laws of logarithms

In a nutshell

Similar to the laws for exponentials, e.g. $$a^x \times a^y = a^{x + y}$$, laws for logarithms exist which you can use to manipulate expressions they feature in. Here you will derive and cover those laws.

Multiplication, division and power laws

Multiplication law

Let $$a$$ be a positive constant not equal to $$1$$​, and let $$m$$​ and $$n$$​ be any real values. Suppose that $$a^m = x$$​, and $$a^n = y$$​. Then the laws of exponentials tell you that:

​$$\begin{aligned}xy &= a^m \times a^n\\&= a^{m+n}\end{aligned}$$​

Applying $$\log_a()$$​ to $$x$$​, $$y$$, and $$xy$$​, you have that:

​$$\begin{aligned}m &= \log_a(x)\\n &= \log_a(y)\\m+n &= \log_a(xy)\end{aligned}$$​​

Put these together to derive the multiplication law for logarithms:

​$$\boxed{\log_a(xy) = \log_a(x) + \log_a(y)}$$​​

Division law

If $$x$$ and $$y$$ are constants with $$y \neq 0$$, then you have that $$\dfrac{x}{y} \times y = x$$. Apply $$\log_a()$$ to this equation, use the multiplication law:

​$$\begin{aligned}\log_a\left(\dfrac{x}{y} \times y \right) &= \log_a(x)\\\log_a\left(\dfrac{x}{y}\right) + \log_a(y) &= \log_a(x)\end{aligned}$$​​

Rearrange this to derive the division law for logarithms:

​$$\boxed{\log_a\left(\dfrac{x}{y}\right)= \log_a(x) - \log_a(y)}$$​​

Power law

You have that $$x = a^m$$. Laws of exponentials tell you that:

​$$\begin{aligned}x^k &= (a^m)^k\\&= a^{km}\end{aligned}$$​​

Apply $$\log_a()$$​ to get:

​$$\log_a(x^k) = km$$​​

Substitute $$m = \log_a(x)$$​ to derive the power law for logarithms:

​$$\boxed{\log_a(x^k) = k\log_a(x)}$$​​

Example 1

Let $$a$$​ be a positive constant not equal to $$1$$​, and let $$x \ge 0$$​. Show that $$\log_a(\sqrt{x}) = \dfrac{\log_a(x)}{2}$$.

You have that $$\sqrt{x} = x^{\frac12}$$. The power law for logarithms tells you that:

​$$\begin{aligned}\sqrt{x} &= x^{\frac12}\\\log_a(\sqrt{x}) &= \log_a(x^\frac12)\\\log_a(\sqrt{x}) &= \dfrac12\log_a(x)\end{aligned}$$​​

Therefore, $$\underline{\log_a(\sqrt{x}) = \dfrac{\log_a(x)}{2}}$$.

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Special cases

For a positive constant $$a$$​ not equal to $$1$$​, recall that the unique value of $$x$$​ such that $$a^x = 1$$​ is $$x = 0$$. Similarly, the unique value of $$x$$​ for which $$a^x = a$$​ is $$x = 1$$​. Therefore, you have the following special cases:

​$$\begin{aligned}\log_a(1) &= 0\\\log_a(a) &= 1\end{aligned}$$​​

 Using this and the division law, you can obtain the following useful case:

​$$\begin{aligned}\log_a\left(\dfrac1x\right) & = \log_a(1) - \log_a(x)\\&= -\log_a(x)\end{aligned}$$​​

Example 2

WIthout using a calculator, find the exact value of $$\log_3\left(\dfrac{\sqrt{27}}{9}\right)$$.

You have that $$\sqrt{27} = 3^{\frac32}$$, and $$9 = 3^2$$. Use the power and division laws to get:

$$\begin{aligned}\log_3\left(\dfrac{\sqrt{27}}{9}\right)&= \log_3(\sqrt{27}) - \log_3(9)\\&= \log_3(3^{\frac32}) - \log_3(3^2)\\&= \dfrac32\log_3(3) - 2\log_3(3)\\&= \dfrac{3}{2} - 2\\&= -\dfrac12\end{aligned}$$​​

Therefore, the exact value of $$\log_3\left(\dfrac{\sqrt{27}}{9}\right)$$ is $$\underline{-\dfrac12}$$.

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