Changing bases

Changing bases

Excerpt: In this resource students explore how logarithms of different bases are related. A set of graphical problems are presented, based on a graph of $\log_5 x$, and a set of algebraic problems.

Warm-up

For any positive number $x$, can you explain why this result is true?

$$2\log_4 x=\log_2 x$$

Can you describe what $\log_4 x$ and $\log_2 x$ are in words?

Suggestion

$\log_4 x$ is the power that we’d need to raise $4$ to if we wanted to get $x$, and $\log_2 x$ is the power that we’d need to raise $2$ to if we wanted to get $x$.

Which of the following is correct?

$$2\log_6 x=\log_3 x \quad 3\log_9 x=\log_3 x \quad 2\log_9 x=\log_3 x$$

Test your answer

Thinking about how logarithms are defined, the equation $2\log_4 x=\log_2 x$ seems to hold because $4$ is $2^2$, rather than because $4$ is $2$ ‘doubled’. Therefore of the three alternatives, $2\log_9 x=\log_3 x$ is likely to be the equivalent result, but for base $3$ and $9$. To check this, let $\log_{9} x = k.$ Then $x=9^{k}$, so $\log_{3}x= \log_{3}9^{k}=\log_{3}3^{2k}=2k= 2\log_9 x.$

What is a general statement of this result?

Suggestion

There are various relationships here that we can start to generalise.

• How is $\log_{a}x$ related to $\log_{a^{2}}x$ ?

• How is $\log_{a}x$ related to $\log_{a^{n}}x$?

• How is $\log_{a}x$ related to $\log_{b}x$? How can we express the relationship between $a$ and $b$?

Test your answer

From the examples above, we can see that $\log_{a}x= 2\log_{a^{2}}x,$ which we could generalise to $\log_{a}x= n\log_{a^{n}}x.$ Thinking about $n$’s role in this equation brings us to

$\log_a x=(\log_a b)(\log_b x)$ or equivalently $\log_{b}x=\frac{\log_{a}x}{\log_{a}b}.$

What conditions must $x,a,b$ satisfy here? Could the second of these ever risk division by $0$?

Some graphical problems

Here is a graph of $y=\log_{5} x$

Use this graph of $y=\log_{5} x$ to help you to draw graphs of

1. $\log_{25}x$

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1. $\log_{\sqrt{5}}x$

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1. $\log_{\frac{1}{5}}x$

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1. $\log_{10} x$

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1. $\log_{x} 5$

Try to make your graphs as accurate as you can, but use the graph of $y=\log_{5}x$ to help you, rather than a calculator.

You could try to sketch the graphs first, thinking about which points the graphs go through and which graphs will give larger or smaller values for the same value of $x$. You may find it helpful to look back at the warm-up.

You can draw your graphs on the printable larger version of the $y=\log_{5}x$ graph.

Graph solutions

Here are graphs 1 and 2 drawn using values from the $y=\log_{5}x$. The relationships between each graph and the graph of $y=\log_{5}x$ are described below.

1. $\log_{25} x = (\log_{25} 5)(\log_{5} x) = \frac{1}{2}\log_{5} x$, so the graph of $y = \log_{25} x$ is a stretch of the graph $y=\log_{5}x$ parallel to the $y$-axis and scale factor $\frac{1}{2}$.

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1. $\log_{\sqrt{5}}x= (\log_{\sqrt{5}} 5)(\log_{5} x)=2\log_{5} x$, so the graph of $y=\log_{\sqrt{5}}x$ is a stretch of the graph $y=\log_{5}x$ parallel to the $y$-axis and scale factor $2$.

Thinking about the definition of logarithms, as we did in the warm-up, these results make sense. For example, we’d expect the power that we’d raise $5$ to in order to get a given value of $x$ to be twice as much as the power we’d need to raise $25$ to.

1. As $\dfrac{1}{5} = 5^{-1}$, the power that we’d raise $\dfrac{1}{5}$ to in order to get a given value of $x$ is $-1$ times the power we’d need to raise $5$ to. Alternatively, using the idea from the warm-up, $\log_{\frac{1}{5}}x=(\log_{\frac{1}{5}} 5)(\log_{5}x)=-\log_{5}x$ and so the graph of $y = \log_{\frac{1}{5}}x$ is a reflection in the $x$-axis of the graph of $y=\log_{5}x$.

2. More work is needed for the graph of $y=\log_{10} x$. We know that $\log_{10}x= \dfrac{\log_{5} x}{\log_{5} 10}$, so we need to divide by $\log_{5}10$ to get the graph of $y=\log_{10}x$. The given graph doesn’t go up as far as $x=10$, but $\log_{5}10 = \log_{5}5+ \log_{5}2 = 1+ \log_{5}2$. From the graph, $\log_{5} 2 \approx 0.42$. Now, $1 \div 1.42 = \frac{100}{142}=\frac{50}{71}\approx\frac{49}{70} = 0.7$ so the graph of $y=\log_{10}x$ is a stretch of $y=\log_{5}x$ parallel to the $y$-axis, scale factor $0.7$.

3. For $x>0$ with $x\neq 1$, ($\log_{5}x)(\log_{x}5)= \log_{5}5=1$, so $\log_{x}5$ is the reciprocal of $y=\log_{5}x$. We can use this to plot particular points to get an idea of the shape of the graph $y=\log_{x}5$.

Alternatively, we can use the behaviour of the function $\log_{5} x$ to sketch the behaviour. For $x>1$, as $x$ increases, $\log_{5}x$ increases and is positive, so $\log_{x}5$ decreases and is positive. For $x$ close to $1$, $\log_{5} x$ is close to $0$, and the graph of $y=\log_{x}5$ has an asymptote at $x=1$. For $0< x <1$, $\log_{5}x$ is negative. It increases to $0$ as $x$ increases to $1$, and tends to $-\infty$ as $x$ decreases to $0$. The graph of $y=\log_{x}5$ gets steeper and closer to the $y$-axis as $x$ decreases to $0$, but it doesn’t actually meet the $y$-axis because $\log_{x}5$ isn’t defined for $x=0$.

Some algebraic problems

1. Evaluate the following. Can you do this without calculating each factor separately? $(\log_3 9)(\log_9 81)$ $(\log_3 81)(\log_{81} 9)$

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1. Write the following as a single logarithm. What conditions must $x$ and $y$ satisfy? $(\log_{10} 5)(\log_{5} x)$ $(\log_{10} x)(\log_{x} y)(\log_{y} 3)$

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1. Simplify the following. What conditions must $x$ and $y$ satisfy? $(\log_{x} y)(\log_{y} 10)(\log_{8} x)$ $(\log_{x} y)(\log_{8} x)(\log_{y} 8)$

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1. The value of $\log_2 3$ is approximately $1.6$. Can you use this to approximate $\log_4 27$? What about $\log_8 27?$ Can you generalise this result?

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1. Evaluate $\dfrac{\log_{2} 5}{\log_{2} 25}$ $\dfrac{\log_{10} 5}{\log_{10} 3}$

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1. Given that $\log_x y=4$ and $\log_y 2 = 9$, evaluate $(\log_x 2)(\log_y x)$ $\dfrac{\log_x 2}{\log_y x}$

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1. What is $(\log_a b^3)(\log_b a^3)$?

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1. What is the relationship between $\log_a b$ and $\log_b a$?

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1. Solve $\log_{x} 7 = \log_{7} x$.

Algebra solutions

1. Evaluate the following. Can you do this without calculating each factor separately? $(\log_3 9)(\log_9 81)$ $(\log_3 81)(\log_{81} 9)$

Using the generalised version of the result in the warm-up,

1. $(\log_3 9)(\log_9 81)=\log_{3}81=4$

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1. $(\log_3 81)(\log_{81} 9)= \log_{3}9=2$.

2. Write the following as a single logarithm. What conditions must $x$ and $y$ satisfy? $(\log_{10} 5)(\log_{5} x)$ $(\log_{10} x)(\log_{x} y)(\log_{y} 3)$

Using the same idea as in question 1, (a) $(\log_{10} 5)(\log_{5} x)=\log_{10} x$.

For part (b), one approach is

$$(\log_{10} x)(\log_{x} y)(\log_{y} 3) = (\log_{10} y)(\log_{y} 3) = \log_{10}3$$

Since $x$ and $y$ both occur as bases of logarithms in these expressions, $x$ and $y$ must both be positive, but can’t be $1$.

1. Simplify the following. What conditions must $x$ and $y$ satisfy? $(\log_{x} y)(\log_{y} 10)(\log_{8} x)$ $(\log_{x} y)(\log_{8} x)(\log_{y} 8)$

2. $(\log_{x} y)(\log_{y} 10)(\log_{8} x)=(\log_{x} 10)(\log_{8} x)= (\log_{8} x)(\log_{x} 10)= \log_{8}10$.

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1. $(\log_{x} y)(\log_{8} x)(\log_{y} 8)= (\log_{8} x)(\log_{x} y)(\log_{y} 8)= (\log_{8} y)(\log_{y} 8) = \log_{8}8=1$.

Again, since $x$ and $y$ both occur as bases of logarithms in these expressions, $x$ and $y$ must both be positive, but can’t be $1$.

1. The value of $\log_2 3$ is approximately $1.6$. Can you use this to approximate $\log_4 27$? What about $\log_8 27?$

If we start by using the law for manipulating logs of powers, we get $\log_{4} 27 = 3 \log_{4} 3$.

From the definition of logarithms, we know that $\log_{4}3$ is smaller than $\log_{2} 3$, so we can expect the answer to be less than $4.8$. In fact, we can say more than this, because $\log_{4} 3< 1 < \log_{2} 3$, so we should expect the answer to be less than $3$.

From the warm-up we know that $\log_{4} 3 = \frac{1}{2}\log_{2} 3$. Therefore $\log_{4} 27 = 3 (\log_{4} 2)(\log_{2}3) \approx 3 \times \frac{1}{2}\times 1.6 = 2.4$.

We know that $\log_{8} 27$ will be smaller than $\log_4 27$. By manipulating powers and changing the base, we get $\log_{8} 27 = 3(\log_{8}2)(\log_{2}3)= \log_{2}3 \approx 1.6$.

Note that from the second part we have $\log_{2^{3}}3^{3}= \log_{2} 3$. In general, $\log_{a^{k}}b^{k}= \log_{a} b$ for any positive $a$, $b$ with $a\neq 1$ and any power $k$. How could you prove this?

1. Evaluate $\dfrac{\log_{2} 5}{\log_{2} 25}$ $\dfrac{\log_{10} 5}{\log_{10} 3}$

2. $\dfrac{\log_{2} 5}{\log_{2} 25}= \dfrac{\log_{2} 5}{(\log_{2}5)(\log_{5}25)}= \dfrac{1}{2}$

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1. $\dfrac{\log_{10} 5}{\log_{10} 3}=\dfrac{(\log_{10} 3)(\log_{3}5)}{\log_{10} 3}=\log_{3}5$

Alternatively, we could use the fraction form of the general result in the warm-up.

1. Given that $\log_x y=4$ and $\log_y 2 = 9$, evaluate $(\log_x 2)(\log_y x)$ $\dfrac{\log_x 2}{\log_y x}$

2. $(\log_x 2)(\log_y x)= (\log_y x)(\log_x 2)= \log_{y} 2 = 9$.

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1. Using ideas from question 5, or the two forms for changing base given in the warm-up, the numerator can be rewritten as $(\log_x y)(\log_{y}2)$ and the denominator as $\dfrac{\log_{x}x}{\log_x y}$.

Therefore

$$\begin{align*} \frac{\log_x 2}{\log_y x} &{}= \frac{(\log_x y)(\log_{y}2)}{\dfrac{\log_{x}x}{\log_x y}}\\ &{}= (\log_{x}y)^{2}\log_{y}2\\ &{}= 16 \times 9\\ &{}= 144 \end{align*}$$

1. What is $(\log_a b^3)(\log_b a^3)$?

Using laws for manipulating logs of powers, $(\log_a b^3)(\log_b a^3)=9(\log_a b)(\log_b a) = 9 \log_{a} a = 9.$

1. What is the relationship between $\log_a b$ and $\log_b a$?

In questions 3 and 7 we had something of the form $(\log_{a} b)(\log_{b} a) = \log_{a} a$, so $(\log_{a} b)(\log_{b} a)=1$, or equivalently $\log_{a} b = \dfrac{1}{\log_{b}a}$. Is there any danger of dividing by $0$ here?

1. Solve $\log_{x} 7 = \log_{7} x$.

Using the relationship in question 8, solving $\log_{x} 7 = \log_{7} x$ is equivalent to solving $(\log_{7}x)^{2} = 1$. Therefore we need to solve $\log_{7} x = 1$ and $\log_{7} x = -1$, which give us $x=7$ and $x=\dfrac{1}{7}$ as solutions.