Academic Skills Advice
Using Logs to Solve Equations with Powers of 𝒙
You might be asked to solve an equation with an unknown power (e.g. 3𝑥 = 243). For this
example you could probably use trial and error and work out that 𝑥 = 5. However the
equation may not be so straightforward, in which case you would use logs to solve it.
You will need to remember the 3rd rule of logs discussed in the last lesson:
𝒍𝒐𝒈(𝑨𝒏 ) = 𝒏𝒍𝒐𝒈(𝑨)
This rule will bring 𝑥 down from the power so that you can solve the equation.
Examples:

Solve the equation: 𝟒𝟑𝒙+𝟐 = 𝟕
Step 1: Take the log of both sides:
𝑙𝑜𝑔(43𝑥+2 ) = 𝑙𝑜𝑔(7)
Step 2: use the rule to bring the power to the front:
(3𝑥 + 2)𝑙𝑜𝑔4 = 𝑙𝑜𝑔7
Step 3: Tidy up and solve the equation as normal:
(3𝑥 + 2) =
𝑙𝑜𝑔7
log 4
(3𝑥 + 2) = 1.404
3𝑥 = 1.404 − 2
3𝑥 = −0.596
𝒙 = −𝟎. 𝟏𝟗𝟗
(N.b. Engineers often use the natural log (𝑙𝑛) instead of log, the method is exactly the
same. Try the above example using 𝑙𝑛 and check that you get the same answer.

𝟐
Solve the equation: 𝟖𝒙 −𝟕 = 𝟏𝟐
Step 1: Take logs of both sides:
Step 2: Bring the power to the front:
Step 3: Tidy up and solve as normal:
𝑙𝑜𝑔(8𝑥
2 −7
) = 𝑙𝑜𝑔(12)
(𝑥 2 − 7)𝑙𝑜𝑔8 = 𝑙𝑜𝑔12
(𝑥 2 − 7) =
𝑙𝑜𝑔12
𝑙𝑜𝑔8
(𝑥 2 − 7) = 1.195
𝑥 2 = 1.197 + 7
𝑥 2 = 8.197
𝒙 = ±𝟐. 𝟖𝟔
© H Jackson 2011 / ACADEMIC SKILLS
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
Solve the equation: (𝟕𝟑𝒙+𝟒 )(𝟕𝒙−𝟑 ) = 𝟖
This question needs an extra step. You need to remember the rules of indices that states
that if you multiply numbers with the same base then you add the powers.
Step 0: Using the rules of indices (add the powers):
Step 1: Take logs of both sides:
(73𝑥+4+𝑥−3 ) = 8
4𝑥+1
(7
)=8
𝑙𝑜𝑔(74𝑥+1 ) = 𝑙𝑜𝑔(8)
Step 2: Bring the power to the front:
(4𝑥 + 1)𝑙𝑜𝑔7 = 𝑙𝑜𝑔8
Step 3: Tidy up and solve as normal:
(4𝑥 + 1) =
𝑙𝑜𝑔8
𝑙𝑜𝑔7
(4𝑥 + 1) = 1.0686
4𝑥 = 1.0686 − 1
4𝑥 = 0.0686
𝒙 = 𝟎. 𝟎𝟏𝟕𝟏𝟓
Remember that for all of the above equations you can check your answer by substituting it
back in the original equation to see if it works.
For example:

(𝟕𝟑(𝟎.𝟎𝟏𝟕𝟏𝟓)+𝟒 )(𝟕𝟎.𝟎𝟏𝟕𝟏𝟓−𝟑 ) = 𝟖
(Try this on your calculator and you’ll see that the value of 𝑥 is correct.)

Find 𝒙 and 𝒚 when: 𝟓𝟑𝒙−𝟒𝒚 = 𝟐𝟓 and
𝟔−𝟐𝒙+𝟒𝒚 = 𝟏
When asked to solve equations with 2 unknown values (in this case 𝑥 and 𝑦) we will
consider each equation separately at 1st and once we have moved the unknowns from the
power (using logs) we will solve them simultaneously.
Equation 1:
Step 1: Take logs of both sides:
Step 2: Bring the power to the front:
Step 3: Tidy up:
𝑙𝑜𝑔(53𝑥−4𝑦 ) = 𝑙𝑜𝑔(25)
(3𝑥 − 4𝑦)𝑙𝑜𝑔5 = 𝑙𝑜𝑔25
3𝑥 − 4𝑦 = 2
Equation 2:
Step 1: Take logs of both sides:
Step 2: Bring the power to the front:
Step 3: Tidy up:
© H Jackson 2011 / ACADEMIC SKILLS
𝑙𝑜𝑔(6−2𝑥+4𝑦 ) = 𝑙𝑜𝑔(1)
(−2𝑥 + 4𝑦)𝑙𝑜𝑔6 = 𝑙𝑜𝑔1
−2𝑥 + 4𝑦 = 0
2
Now we have 2 equations that can be solved simultaneously (see lecture 4 for more
information on solving simultaneous equations):
3𝑥 − 4𝑦 = 2
−2𝑥 + 4𝑦 = 0
𝒙=𝟐
Add the equations to get:
Substitute in equation 1:
3(2) − 4𝑦 = 2
6 − 2 = 4𝑦
𝒚=𝟏
We have found that the values of 𝑥 = 2 and 𝑦 = 1 will work in both equations. Use your
calculator to check that the values are correct.
An alternative method:
If you are happy with the above method for solving this type of equation you don’t need to
read the following. It’s just included for completeness as some people like to think about it
this way.
Remember what a log means:
e.g.
𝑙𝑜𝑔3 3 means “what power do I need to raise 3 to, to get 3?” (answer =1)
So if the base and the number are the same then the answer will always be 1.
In general:
𝒍𝒐𝒈𝒏 𝒏 = 𝟏
We could use this fact to solve this type of equation.

Solve the equation: 𝟒𝟑𝒙+𝟐 = 𝟕
(Use this number as the base.)
Step 1: Take the log of both sides using 4 as the base:
𝑙𝑜𝑔4 43𝑥+2 = 𝑙𝑜𝑔4 7
Step 2: use the log rule to bring the power to the front:
(3𝑥 + 2)𝑙𝑜𝑔4 4 = 𝑙𝑜𝑔4 7
Step 3: we know that 𝑙𝑜𝑔4 4 = 1 so we now have:
Step 4: Tidy up and solve the equation as normal.
© H Jackson 2011 / ACADEMIC SKILLS
(3𝑥 + 2)(1) = 𝑙𝑜𝑔4 7
(3𝑥 + 2) = 1.404
3𝑥 = 1.404 − 2
3𝑥 = −0.596
𝒙 = −𝟎. 𝟏𝟗𝟗
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