Part 7: Year 10 Logarithms

In this article, we take the mystery out of logarithms.

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Year 10 Logarithms

In this article, we give you an overview of Year 10 Logarithms. Logarithims are used to calculate how loud something is, how acidic it might be, or how violent an earthquake is. Logarithms have important real world applications and will be an important element of Year 11 and 12 Maths.

 

Outline of Year 10 Logarithms

Being able to understand and work with logarithms effectively is an important skill, especially since they form the basis of practical scales such as the Richter scale, which is used to measure the magnitude or severity of an earthquake.

This page will give examples of how to simplify logarithmic expressions using logarithmic laws, as well as an outline of the change of base formula for logarithms.

 

NSW Syllabus Outcomes

Below are the NESA expecations for Logarithms

Stage 5.3: Use the definition of a logarithm to establish and apply the laws of logarithms (ACMNA265)

  • Define ‘logarithm’: the logarithm of a number to any positive base is the index when the number is expressed as a power of the base,
    ie. \(a^{x}=y⇔\log_a y=x\), where \(a>0,y>0\).
  • Deduce the following laws of logarithms from the laws of indices:
    • \(\log_a x+\log_a y=\log_{a} (xy)\)
    • \(\log_a x-\log_a y=\log_a (\frac{x}{y})\)
    • \(\log_a x^n= n \log_a x\)
  • Apply the laws of logarithms to simplify simple expressions,
    eg. \(\log_{2}⁡8, \ \log_{81}⁡3, \ \log_{10⁡}25+\log_{10}⁡4, \ 3 \log_{10}⁡2+\log_{10⁡}(12.5), \ \log_2⁡18-2 \log_2⁡3\)
  • Simplify expressions using the laws of logarithms,
    eg. simplify  \(5 \log_a⁡ a – \log_{a}⁡ a^{4}\)

 

Solve simple exponential equations (ACMNA270)

  • solve simple equations that involve exponents or logarithms,
    eg.  \(2^{t}=8, \ 4^{(t+1)}=\frac{1}{(8\sqrt2)}, \ \log_{27⁡}3=x, \ \log_{4}⁡x=-2\)

This looks pretty intimidating, but it shouldn’t be. All you’re doing here is learning what logarithms are. Similar to index laws, once you learn what logarithms are you will learn the laws that govern them and use them to solve equations.

 

Assumed knowledge

Students should be familiar with the definition of a logarithm and how to interchange between expressions involving logarithms and indices.

We will also assume knowledge of the following result:

\(\log_{a⁡}a=1\)

 

In this article

We will discuss:

  • Logarithms of products and quotients
  • Logarithm of power
  • Change of base theorem
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Year 10 Logarithms

Logarithms allow us to re-express equations of the form \(a^y=x\)  as  \(log_a⁡x=y\), where \(a>0\) and \(x>0\).

We read this as ‘log to the base \(a\) of \(x\) is \(y\)‘.

For example, we can re-express \(2^3=8\) as  \(log_2⁡8=3\).

 

Now, recall how we have the following laws for indices:

  • \(b^m×b^n=b^{m+n}\)
  • \(b^m÷b^n=b^{m-n}\)

These indicial laws can be used to derive laws for logarithms as below.

 

Logarithms of products and quotients

We won’t go through the proofs here, but feel free to try them by yourself using the index laws listed above!

  • \(log_{b⁡}(xy)= \log_{b}⁡x+\log_{b}⁡y\)
  • \(log_{b}(x÷y)= \log_{b} \frac{x}{y}=log_{b}⁡x-\log_{b}⁡y\)

You need to use these logarithmic laws when you’re asked to express the sum of logarithms as one single logarithm.

Make sure that when you’re doing this, the logarithms you’re combining all have the same base (notice that all terms in the equations above are to the base).

 

Example

Express the following as a single logarithm:

  1. \(\log_{5⁡}10+\log_{5}⁡7+log_{5}⁡2 \)
  2. \(\log_{2⁡}(3y)+\log_2(22y)-\log_2⁡{11}\)

 

Solution

1. Since each of the logarithms is to the base \(5\), we can apply the logarithmic law \(\log_{5}⁡x+log_{5}⁡y=log_{5}(xy)\) to combine them.

Applying this to the first two terms gives us

\(\log_{5}⁡10+\log_{5}⁡7=\log_{5}(10×7)=log_{5}⁡70\)

 

Our whole expression then simplifies to \(log_{5}⁡70+\log_{5}⁡2\).

We can apply the same logarithmic law to combine these two logarithms as \(\log_{5}(70×2)=\log_{5}⁡140\).

Of course, we could also have combined the three logarithms in one go:

\(log_{5}⁡10+\log_{5}⁡7+\log_{5}⁡2=\log_5(10×7×2)=\log_{5}⁡140\)

 

2. Here we have pronumerals in the logarithms, but don’t worry as the logarithmic laws still apply!

We can combine \(log_2⁡(3y)+log_2⁡(22y)\) as \(\log_2⁡(3y×22y)=log_2⁡(66y^2 )\).

Our expression then reduces down to \(log_2⁡(66y^2 )-log_2⁡(11y)\).

Applying the logarithmic law \(\log_2⁡x-\log_2⁡y=\log_2⁡\frac{x}{y}\) then allows us to simplify

\(\log_2⁡(66y^2 )-\log_2⁡(11y)\) as \(\log_2⁡\frac{66y^2}{11y}=\log_2⁡(6y)\).

 

You might also be asked some questions that require you to use these logarithmic laws the other way around.

That is, instead of combining many logarithms into a single one, you might need to split a logarithm apart into separate logarithms.

 

Example

Use the following values \(\log_{3}⁡2≈0.631\) and \(\log_{3}⁡5≈1.465\) to evaluate \(\log_{3}⁡10\).

 

Solution

Basically, the aim is to express \(\log_{3}⁡10\) in terms of \(\log_{3}⁡2\) and \(\log_{3}⁡5\) so that we can use the approximations given.

We can apply the logarithmic laws to write \(\log_{3⁡}10=\log_{3⁡}(2×5)=\log_{3}⁡2+\log_{3}⁡5\).

The approximations then give us \(\log_{3⁡}10≈0.631+1.465=2.096\).

 

Logarithm of a power

Recall the index law:

\((b^m )^n=b^{mn}\).

From this it is possible to derive the following logarithmic law (prove it if you’re keen!):

\(log_{b}(x^n)=n \log_{b}⁡x\).

This law can be used to remove the power inside the logarithm, taking it to the front instead.

 

Example

1. Expand \(\log_{7⁡}8^2 \).

2. Expand \(log_{3⁡}(\frac{x^3 \sqrt{y}}{z})\).

 

Solution

1. We can take the power inside the logarithm to the front. This gives us \(2\log_{7}⁡8\).

2. First, we use the quotient logarithmic law to expand the expression out as \(log_{3}(x^{3} \sqrt{y})-\log_{3}⁡z\).

The product logarithmic law then lets us expand this out as \(\log_{3⁡}(x^3 )+ \log_{3}\sqrt{⁡y}-\log_{3}⁡z=\log_{3}(x^3 )+ \log_3⁡y^{\frac{1}{2}} -log_{3}⁡z\).

Finally, applying the logarithm of a power law gives us \(3 \log_{3⁡}x+ \frac{1}{2} \log_{3}⁡y -log_{3}⁡z\).

 

Change of base theorem

Calculators only let you calculate the logarithm of a number to two bases – specifically, base \(10\) or base \(e\).

The log button on your calculator calculates the logarithm of a number to base \(10\) whilst the \(ln\) button does it for base \(e\).

 

But what if we want to compute the logarithm of a number to a different base?

The change of base theorem allows for this, stating that:

\(\log_{a}⁡n=\frac{\log_{b}⁡n}{\log_{b}⁡a}\)

 

The base \(b\) can be any integer, but make sure that it is the same in the numerator as the denominator.

Usually we use \(b=10\) or \(b=e\) so that we can use our calculator to evaluate the logarithm.

 

Example

1. Evaluate \(\log_{25}⁡11\).

2. Simplify \(\log_{7⁡}11×\log_{3}⁡7\).

3. Solve for \(x\) in the following: \(\log_{25}⁡3+\log_{5}⁡7=\log_5⁡x\).

 

Solutions

1. We can use the change of base formula to rewrite \(\log_{25}⁡11\) in either base \(10\) or base \(e\).

Change of base to base \(10\) gives us \(\log_{25}⁡11=\frac{\log_{10}⁡11}{\log_{10}⁡25} ≈\frac{1.041}{1.398}=0.745\).

If we change to base \(e\), we get \(\log_{25}⁡11=\frac{\log_{e}⁡11}{\log_{e}⁡25} ≈\frac{2.398}{3.219}=0.745\), which is the same result.

 

2. Notice here that \(7\) appears as the base of the first logarithm and as the number inside the second logarithm.

This suggests to us that we should use the change of base formula on the first logarithm. We need to change the base to  \(3\), so that we have a \(\log_{3}⁡7\) on the denominator: \(\log_{7}⁡11=\frac{\log_{3}⁡11}{\log_{3}⁡7}\) .

The denominator then cancels out with the \(\log_{3⁡}7: \log_{7}⁡11× \log_{3}⁡7= \frac{\log_{3}⁡11}{\log_{3⁡}7} × \log_{3⁡}7=\log_{3}⁡11\).

 

3. First, notice that two of the logarithms are to the base \(5\), whilst the other one is to the base \(25\)

In order to solve the equation, we must have all the logarithms in the same base.

Let’s use the change of base formula to convert \(\log_{25⁡}3\) to base \(5: \log_{25}⁡3=\frac{\log_{5}⁡3}{\log_{5}25}\) .

Using the logarithmic power law changes this to \(\frac{\log_{5}⁡3}{\log_{5}25} =\frac{\log_{5}⁡3}{2\log_{5}⁡5 }=\frac{\log_{5}⁡3}{2}\).

 

Our equation then becomes:

\(\frac{\log_{5}⁡3}{2}+\log_{5}⁡7=\log_{5}⁡x\), which is the same as \(\log_{5}⁡3+2 \log_{5}⁡7=2 \log_{5}⁡x\).

We need to combine the left-hand side into a single logarithm so that we can compare with the right-hand side and find out what x is.

Using a combination of logarithmic laws, we get: \(\log_5⁡(3×7^2)=\log_{5}⁡x^2 \),

i.e. \(x^2=3×7^2\) and \(x=±7\sqrt{3}\).

 

But note

This is not the final answer!

One thing you have to be careful about when solving logarithmic equations is that you satisfy any restrictions on \(x\).

Looking back at our original equation, we can see that x appears in the \(\log_5⁡x\) term. Recall that any number inside a logarithm has to be positive \((>0)\).

Therefore, we must have that \(x>0\),

i.e. \(x= 7 \sqrt{3}\).

 

Note to students

Restrictions on solutions to logarithmic equations is just something that you have to keep an eye out for.

It’s a good idea to make a note right at the start of the question what the restrictions on \(x\) are (if any).

Note that if, for instance, you had \(log_{5⁡}(x-3)\) appearing in the equation (instead of \(log_{5⁡}x\)), you would need to check that your final answer satisfies \(x-3>0\), i.e. \(x>3\).

 

Summary

Use the following logarithmic laws to expand and simplify expressions involving logarithms, as well as solve logarithmic equations:

  • \(\log_{b}(xy)=\log_{b}⁡x+\log_{b}⁡y\)
  • \(\log_{b}⁡(x÷y)=\log_b⁡ \frac{x}{y}=\log_{b}⁡x-\log_{b}⁡y\)
  • \(\log_{b⁡}(x^n)=n \log_{b}⁡x\)

 

The change of base theorem is essential if you want to use your calculator to evaluate the logarithm of a number to a base that is not \(10\) or \(e\). It is also useful in simplifying certain logarithmic equations:

  • \(\log_{a}⁡n=\frac{log_{b}⁡n}{log_{b}⁡a}\)

When solving equations with logarithms, make sure to pay attention to any restrictions on solutions for x as a result of expressions with \(x\) appearing inside any of the logarithms in the equation.

 

Checkpoint Questions

Questions

Check your skills with the following 8 exercises!

 

Express the following as a single logarithm:

1. \(log_3⁡15-log_3⁡5+log_3⁡8\)

2. \(3 log_x⁡y+log_x⁡(yz)-log_x⁡(x+1)\)

 

Expand:

3. \(\log_{5}⁡125\)

4. \(log_{10⁡} \frac{w\sqrt[3]{xy}}{z^2}\)

 

Use the following values \(\log_5⁡10≈1.431\) and \(\log_5⁡3≈0.683\) to evaluate:

5. \(\log_{5}⁡2700\)

 

Evaluate:

6.  \(\log_{5}⁡33\)

 

Simplify:

7. \(\log_3⁡7×\log_7⁡3×\log_3⁡9\)

 

Solve for \(x\):

8. \(\log_9⁡(x-1)+\log_3⁡5=\log_3⁡15\)

 

Solutions

1.

\(\log_3⁡\frac{15×8}{5}=log_3⁡24\)

 

2.

\( \log_{x} \frac {y^{3} \times yz}{x+1} = \log_{x} \frac{y^4 z}{x +1}\)

 

3.

\(3 \log_5⁡5=3\)

 

4.

\( 3\log_{10}⁡w+\frac{1}{2}  \log_{10}⁡x+\frac{1}{2} \log_{10}⁡y-2 log_{10}⁡z\)

 

5.

\(\log_5⁡(3^3×10^2)=3 \log_5⁡3+2 log_5⁡10≈3×0.683+2×1.431=4.911\)

 

6.

\(\frac{\log_{10}⁡33}{\log_{10}⁡5} ≈\frac{1.519}{0.699}≈2.173\)

 

7.

\(\frac{\log_7⁡7}{log_7⁡3} ×\log_7⁡3×2 \log_3⁡3=1×2=2\)

 

8.

\(
\frac{\log_3⁡(x-1)}{log_3⁡9} =\log_{3}⁡15-log_{3}⁡5 \\
\frac{\log_3⁡(x-1)}{2}=log_{3⁡}3 \\
\log_3(x-1) = 2\\
x-1 = 3^2\\
x = 9+1\\
x = 10 \\
\)

 

Thanks!

Thanks for taking the time to read our Beginner’s Guide to Year 10 Maths. Year 10 Maths is really quite important and we want you to achieve your best.

We hope that you’ve learnt something new from this subject guide, so now you can get out there and ace Mathematics!

© Matrix Education and www.matrix.edu.au, 2019. Unauthorised use and/or duplication of this material without express and written permission from this site’s author and/or owner is strictly prohibited. Excerpts and links may be used, provided that full and clear credit is given to Matrix Education and www.matrix.edu.au with appropriate and specific direction to the original content.

 

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