Further Logarithms

The logarithm $\left(\log\right)$ of a number $x$ to base $b$ is the power to which $b$ must be raised to get the number $x$ .
$\textcolor{#287FD4}{b^a}=\textcolor{#B53329}{x}$ $\log_{\color{#287FD4}b}(\textcolor{#B53329}{x})=\textcolor{#287FD4}{a}$

NOTE: A dot ( $\cdot$ ) is used to represent multiplication throughout this article, to avoid confusing a multiplication symbol with the letter $x$ .

There are many different rules that are used to manipulate logarithms and achieve a desired result. There are a few main rules that are required for KCSE, all of which will be covered in this article. Note that you do not need to know the names of these rules, only how to apply them. You may briefly review the laws of indices in order to better understand the reasoning behind the various logarithm rules covered in this topic.

Logarithm Rules

Product Rule: $\log_b(MN)=\log_b(M)+\log_b(N)$

Why is this true?

Let $M=b^x$ and $N=b^y$ . Thus, it is also true that $\log_b(M)=x$ and $\log_b(N)=y$ . Now we have:
$\begin{align*}\log_b(MN)&=\log_b(b^x\cdot b^y) \\[6pt] &=\log_b(b^{x+y}) \\[6pt] &= x+y\tag*{$\log_b(b^c)=c$} \\[6pt] &= \log_b(M) + \log_b(N) \end{align*}$

Quotient Rule: $\log_b \left(\frac{M}{N}\right)=\log_b(M)-\log_b(N)$

Where does this come from?

The proof of this property follows a method similar to the one used above.

Again, if we let M, equals, b, start superscript, x, end superscript and N, equals, b, start superscript, y, end superscript, then it follows that log, start base, b, end base, left parenthesis, M, right parenthesis, equals, x and log, start base, b, end base, left parenthesis, N, right parenthesis, equals, y.

We can now prove the quotient rule as follows:

$\begin{aligned}\log_b\left(\dfrac{M}{N}\right)&=\log_b\left(\dfrac{b^x}{ b^y}\right)&&\small{\gray{\text{Substitution}}}\\ \\ &=\log_b(b^{x-y})&&\small{\gray{\text{Properties of exponents}}}\\ \\ &=x-y&&\small{\gray{\text{$\log_b(b^c)=c$}}}\\ \\ &=\log_b(M)-\log_b(N)&&\small{\gray{\text{Substitution}}} \end{aligned}$

Power Rule: $\log_b(M^p)=p\log_b(M)$

Proof

Let

M=b^x

. From this, we have that

\log_{b}(M)=x

. Now, we perform a series of operations:

\begin{align*} \log_{b}(M^p)&=\log_{b}((b^x)^p) \\[6pt] &=\log_{b}(b^{xp}) \\[6pt] &=xp \\[6pt] &=\log_{b}(M)\cdot p \tag*{$x=\log_{b}(M)$} \\[6pt] &=p\cdot \log_{b}(M) \end{align*}

Change of Base Rule: $\log_{b}(a)=\frac{\log_{x}(a)}{\log_{x}(b)}$

This rule is especially useful when dealing with bases that are not base 10, as most calculators can only calculate logarithms in base 10.

Why does this work?

Let’s start with a concrete example. Using the above formula, we want to show that $\displaystyle \log_{2}(50)=\frac{\log(50)}{\log(2)}$ . Let’s use $n$ as a placeholder for $\log_{2}(50)$ . In other words, $\log_{2}(50)=n$ . From the definition of logarithms, we see that $2^n=50$ . Now, we can perform a sequence of operations on this equality:

$2^n=50$ $\tag*{If A = B, then log(A) = log(B)} \log(2^n)=\log(50)$ $\tag*{Power Rule} n\log(2)=\log(50)$ $\tag*{Divide both sides by log(2)}n=\frac{\log(50)}{\log(2)}$ Since $n$ was defined as $\log_{2}(50)$ , we have that $\displaystyle \log_{2}(50)=\frac{\log(50)}{\log(2)}$ , as desired!

We can now generalise this to prove the change of base rule. We can do this by simply changing 2 to $b$ , 50 to $a$ , and the new base to $x$ . Let $\log_{b}(a)=n$ . We can rewrite this using the definition of logarithms as $b^n=a$ . Now we have: $b^n=a$ $\log_{x}(b^n)=\log_{x}(a)$ $n\log_{x}(b)=\log_{x}(a)$ $n= \frac{\log_{x}(a)}{\log_{x}(b)}$ Since $n=\log_{b}(a)$ , we have that: $\displaystyle\log_{b}(a)=\frac{\log_{x}(a)}{\log_{x}(b)}$ .

Log of 1: $\log_{b}(1) = 0$

Log of a number to its own base: $\log_{b}(b) = 1$

Log of a power of a number to its own base: $\log_{b}(b^c) = c$

These are the basic logarithm rules. Their main application in 844 math is evaluating expressions that require you to find the power that a number is raised to. But in the real world, logarithms are used to find compound interest, model exponential growth and decay, to find the pH level of a substance, and even determine the magnitude of an earthquake. Now that you know the basics, let’s move on to some examples.

Example 1

Find the value of $x$ that satisfies the equation:

$\log(2x-11)-\log(2) = \log(3) – \log(x)$

Let us evaluate the left side of the equation first. We can combine the two logarithms using the quotient rule to get:
$\log\left(\frac{2x-11}{2}\right)$ On the right side, we can use the quotient rule to obtain: $\displaystyle \log\left(\frac{3}{x}\right)$

Now, we have: $\log\left(\frac{2x-11}{2}\right) = \log\left(\frac{3}{x}\right)$ Since these two logarithms are equal, the fractions in the brackets must be equal. Therefore: $\frac{2x-11}{2} = \frac{3}{x}$ $x(2x-11) = 2(3)$ $2x^2-11x = 6$ $2x^2-11x-6=0$ $(2x+1)(x-6)=0$ $x=-\frac{1}{2} \text{ or } 6$

Now this is the key part of the solution. When we get 2 values of x, we must substitute each of them into the original logarithmic equation to see if they both satisfy the equation. Let us start by substituting $x=-\frac{1}{2}$ : $\log\left(2\left(-\frac{1}{2}\right)-11\right)-\log \, 2 =\log\, 3 – \log \left(-\frac{1}{2}\right)$ $\log(-12)-\log \, 2 = \log\, 3 – \log \left(-\frac{1}{2}\right)$

Note that we cannot have the logarithm of a negative number, because no positive number $b ^ a$ gives a negative number. For example, $10^x$ is never a negative number, for any value of $x$ . Therefore, $x=-\frac{1}{2}$ is not a solution to this equation.

Let us try substituting $x = 6$ : $\log(2(6)-11)-\log \, 2 = \log\, 3 – \log (6)$ $\log(1)-\log \, 2 = \log\, 3 – \log (6)$ Since there is no logarithm of a negative number, we see that $\boxed{ x = 6 }$ is a solution.

Example 2

Solve for x in the equation: $(\log_{3}x)^2-\frac{1}{2}\log_{3}x=\frac{3}{2}$

One of the logarithms is raised to the power of 2, and this would be quite difficult to deal with. We need to try to find a way to eliminate the square. We can do this by expressing the equation as a quadratic. Let $\log_{3}x$ be $y$ . We have that $y^2-\frac{1}{2}y-\frac{3}{2}=0$ Multiplying by 2: $2y^2-y-3=0$ $(y+1)(2y-3)=0$ $y = -1 \text{ or }1.5$ Substituting $y = -1$ , $\log_{3}(x)=-1$ $3^{-1}=x$ $x=\frac{1}{3}$ Substituting $y = 1.5$ : $\log_{3}(x)=1.5$ $3^{1.5}=x$ $x=5.196$ Thus, our solutions are $\boxed{ x=\frac{1}{3} }$ and $\boxed{ x=5.196 }$ .

In this example, note that the solution lay in substituting something relatively simple in place of something complex. In this case, we substituted $y$ in place of $\log_{3}x$ . This is true of a lot of problems in math. If you can represent something in a simpler way, the solution becomes clearer. This is the power of algebra.

And that’s it! This topic is relatively short and easy to neglect. However, make sure to gain an intuitive sense of when to use which rule. Remembering each of these rules can help you determine how to manipulate a question to make one of the rules apply.

Further Logarithms

The logarithm \left(\log\right) of a number x to base b is the power to which b must be raised to get the number x .\textcolor{#287FD4}{b^a}=\textcolor{#B53329}{x} \log_{\color{#287FD4}b}(\textcolor{#B53329}{x})=\textcolor{#287FD4}{a}

Logarithm Rules

Product Rule: \log_b(MN)=\log_b(M)+\log_b(N)

Quotient Rule: \log_b \left(\frac{M}{N}\right)=\log_b(M)-\log_b(N)

Power Rule: \log_b(M^p)=p\log_b(M)

Change of Base Rule: \log_{b}(a)=\frac{\log_{x}(a)}{\log_{x}(b)}

Log of 1: \log_{b}(1) = 0

Log of a number to its own base: \log_{b}(b) = 1

Log of a power of a number to its own base: \log_{b}(b^c) = c