# Solving Equations with Logarithms – A Level Maths Revision

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### Where to Start?

Once you have a grip on the basics of logarithms, you need to be capable of solving equations with logarithms.

If you want to check your understanding of logarithms before you start, try this multiple-choice quiz. The content of the blog below, along with a more extensive set of questions, can be downloaded in PDF and PowerPoint form here. Once you’re happy you’ve fully grasped the AS Level content on exponentials and logarithms, try these exam-style questions.

A major part of solving equations is ‘doing the same to both sides’. This can include taking logs of both sides. You can use this information to solve equations with logarithms and exponentials: \begin{aligned} &\log_a{b} = c \text{ can be written as } a^c=b \\ &\text{(where }a>0, a \neq 1 \text{ and }a\text{ is real)} \\ \\ &\log_a{b}+\log_a{c}=\log_a{bc} \\ &\log_a{b}-\log_a{c}=\log_a{\frac{b}{c}} \\ &\log_a{b^c}=c\log_a{b} \end{aligned} \begin{aligned} &\log_a{b} = c \text{ can be written as } a^c=b \\ &\text{(where }a>0, a \neq 1 \text{ and }a\text{ is real)} \\ \\ &\log_a{b}+\log_a{c}=\log_a{bc} \\ &\log_a{b}-\log_a{c}=\log_a{\frac{b}{c}} \\ &\log_a{b^c}=c\log_a{b} \end{aligned}

### Example Question 1 $3^x = 7$

To solve this equation without knowing about logarithms, you would probably use trial and improvement. x must lie between 1 and 2 because 31 = 3 and 32 = 9, so you could use a calculator to evaluate 31.5 and so on.

However, if the unknown in an equation appears as a power, you can get straight to the answer by taking logs of both sides. It’s up to you what base you use, but if you use the base given (3 in this example), the solution will be simpler. $3^x = 7$

Taking logs (to base 3) of each side gives: $\log_3{3^x} = \log_3{7}$

Using the log law for powers gives: $x\log_3{3}=log_3{7}$

The unknown no longer features as a power and log33=1: \begin{aligned} &x = \log_3{7} \\ &x=1.77\text{ (to 2d.p.)} \end{aligned}

When you take logs of both sides, that log can be to any base. Although it can be easier to use a base number from the question, log to the base 10 is often used instead, as it is a single button on your calculator (usually just labelled log) which does not require you to enter a base. This is helpful if the base number in the question is awkward, as even slight rounding in a log question can lead to considerable inaccuracy in your answer. Let’s try the same question using log10: \begin{aligned} &3^x = 7 \\ &\log_{10}{3^x} = \log_{10}{7} \\ & x\log_{10}{3}=log_{10}{7} \end{aligned}

The unknown no longer features as a power. log103 can’t be cancelled out in the way log33 can be. \begin{aligned} &x = \log_{10}{7} \div \log_{10}{3} \\ &x=1.77\text{ (to 2d.p.)} \end{aligned}

As a final alternative: \begin{aligned} &\log_a{b} = c \text{ can be written as } a^c=b \\ &3^x = 7 \text { can be written as } \log_3{7}=x \\ &x=1.77\text{ (to 2d.p.)} \end{aligned}

### Example Question 2 $5^x = 3^{x-1}$

Using the base number given in the equation makes finding a solution easier – here there are two to pick from. Let’s choose base 5. \begin{aligned} &\log_5{5^x} = \log_5{3^{x-1}} \\ & x\log_5{5}=(x-1)\log_5{3} \\ &x = x\log_5{3}-\log_5{3} \\ &\log_5{3} = x\log_5{3} - x \\ &\log_5{3} = x(\log_5{3}-1) \\ &x = \frac{\log_5{3}}{\log_5{3}-1} \\ &x = -2.15 \text{ (to 2d.p.)} \end{aligned}

Again, substitute it back in to check that it gives correct values.

As well as using logs to solve equations involving exponentials, you’ll need to be able to solve equations which already contain logs. You’ll usually use log laws to do so.

### Example Question 3 $\log_3{(x+2)}+\log_3{5}=\log_3{(9x)}$

Using the first law of logs gives: $\log_3{(5x+10)}=\log_3{(9x)}$

Therefore: \begin{aligned} &5x + 10 = 9x \\ &x = 2.5 \end{aligned}

Substitute it back into the original equation to check.

### Example Question 4 $\log_7{(x-2)}+\log_7{x}=\log_7{(x+10)}$

Using the first law of logs gives: $\log_7{(x^2-2x)}=\log_7{(x+10)}$

In this case, removing the logs gives a quadratic. This can be solved as normal: \begin{aligned} & x^2-2x=x + 10 \\ &x^2-3x-10=0 \\ &(x-5)(x+2)=0 \\& x = 5 \text{ or } x=-2 \end{aligned}

Again, substitute it back into the original equation to check. This time, you will find that -2 is not a valid solution because there is no power which acts upon 7 to give a negative result – the log of a negative number gives a mathematical error on a calculator.

### Example Question 5 $\log_4{(x^2+10x)}=2+\log_4{(x+1)}$

You may also see equations with a mix of logarithmic and non-logarithmic terms: \begin{aligned}&\log_4{(x^2+10x)}=2+\log_4{(x+1)} \\& \log_4{(x^2+10x)}-\log_4{(x+1)}=2 \\ & \log_4{\frac{x^2+10x}{x+1}} = 2\end{aligned}

The inverse of taking the logarithm to base 4 of a term is raising 4 to the power of that term. Therefore, if both sides are taken as a power of 4, you get: \begin{aligned} &\frac{x^2+10x}{x+1}=4^2 \\ \\ &x^2 + 10x = 16(x+1) \\ &x^2 + 10x -16x - 16 = 0 \\ &(x-8)(x+2)=0 \\&x = 8 \text{ or } x=-2 \end{aligned}

As before, you can reject x = -2 as a possible answer, leaving x = 8.

## Solving Equations with Logarithms – Practice Questions

1. $4^x = 18$

2. $4^x - 6 = 11$

3. $7^{3x} = 5$

4. $8^{x-1} = 4^x$

5. $4^x \times 4^{x+5} = 4^{x+11}$

6. $3 \times 2^x = 9$

7. $5^{2x-3} = 6^{150}$

8. $\log_4{p} + log_4{(p-8)} = \log_4{(2p-10)} + \log_4{2}$

9. $\log_{10}{s} +\log_{10}{(s-1)} = \log_{10}{(3s + 12)}$

10. $\log_6{(m^2-41m)} = 2 + \log_6{(1-m)}$

1. $\boldsymbol{ 4^x = 18 }$ \begin{aligned} &\log_4{4^x}=\log_4{18} \\&x=\log_4{18}\\&x=2.08 \end{aligned}

2. $\boldsymbol{ 4^x - 6 = 11 }$ \begin{aligned} &4^x = 17 \\ &\log_4{4^x} = \log_4{17} \\ &x\log_4{4}=\log_4{17} \\ & x = \log_4{17} \\ &x=2.04 \end{aligned}

3. $\boldsymbol{ 7^{3x} = 5 }$ \begin{aligned} &\log_7{7^{3x}}=\log_7{5} \\&3x\log_7{7}=\log_7{5} \\ &3x = \log_7{5} \\ &x=0.276 \end{aligned}

4. $\boldsymbol{ 8^{x-1} = 4^x }$ \begin{aligned} &\log_4{8^{x-1}}=\log_4{4^x} \\ &(x-1)\log_4{8}=x\log_4{4} \\ &x\log_4{8}-\log_4{8}=x \\ &\log_4{8} = x\log_4{8} - x \\ &\log_4{8}=x(\log_4{8}-1) \\&x=\frac{\log_4{8}}{\log_4{8}-1} \\ &x=3 \end{aligned}

5. $\boldsymbol{ 4^x \times 4^{x+5} = 4^{x+11} }$ \begin{aligned} &4^{2x+5} = 4^{x+11} \\ &2x + 5 = x + 11 \\&x=6 \end{aligned}

6. $\boldsymbol{ 3 \times 2^x = 9 }$ \begin{aligned} &2^x=3\\&\log_2{2^x}=\log_2{3} \\&x\log_2{2}=\log_2{3} \\&x = \log_2{3} \\ &x=1.58 \end{aligned}

7. $\boldsymbol{ 5^{2x-3} = 6^{150} }$ \begin{aligned} &\log_5{5^{2x-3}}=\log_5{6^{150}} \\ &(2x-3)\log_5{5} = \log_5{6^{150}} \\ &2x-3=\log_5{6^{150}} \end{aligned}

If you now try to use your calculator to evaluate the right-hand side, you’ll find you can’t: 6150 is too large to work with. But, if you rearrange the right-hand side, this can be circumvented: \begin{aligned} &2x-3=150\log_5{6} \\ &x=0.5 \times (150\log_5{6}+3) \\ &x = 85.0 \end{aligned}

8. $\boldsymbol{ \log_4{p} + log_4{(p-8)} = \log_4{(2p-10)} + \log_4{2} }$ \begin{aligned} &log_4(p^2-8p)=\log_4{(4p-20)} \\ &p^2 -8p = 4p-20 \\ &p^2-12p+20=0 \\&(p-10)(p-2) = 0 \\ &p=10 \text{ or } p = 2 \end{aligned}

9. $\boldsymbol{ \log_{10}{s} +\log_{10}{(s-1)} = \log_{10}{(3s + 12)} }$ \begin{aligned} &\log_{10}{(s(s-1))} = \log_{10}{(3s+12)} \\ &s^2-s=3s+12 \\&s^2-4s-12=0 \\ &(s-6)(s+2) = 0 \\ &s=6 \text{ or } s=-2 \text{ (reject)} \end{aligned}

10. $\boldsymbol{ \log_6{(m^2-41m)} = 2 + \log_6{(1-m)} }$ \begin{aligned} &\log_6{(m^2-41m)}-\log_6{(1-m)}=2 \\ & \log_6{\frac{m^2-41m}{1-m}}=2 \\ &6^2=\frac{m^2-41m}{1-m} \\ &36(1-m)=m^2-41m \\ & 36-36m = m^2-41m \\&m^2-5m-36=0\\&(m-9)(m+4)=0 \\&m=9 \text{ or } m=-4 \text{ (reject)} \end{aligned}

Do you feel confident now in solving equations with logarithms? If so, feel free to move on to more of our blogs and practice content here! You can also subscribe to Beyond for access to thousands of secondary teaching resources. You can sign up for a free account here and take a look around at our free resources before you subscribe too.