Using Logs to Solve Equations with Powers of X

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:

logAn=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: log43x+2=log7

Step 2: use the rule to bring the power to the front: 3x+2log4=log7

Step 3: Tidy up and solve the equation as normal: 3x+2=log7log4

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.

· Solve the equation: 8x2-7=12

Step 1: Take logs of both sides: log8x2-7=log12

Step 2: Bring the power to the front: x2-7log8=log12

Step 3: Tidy up and solve as normal: x2-7=log12log8

x2-7=1.195

x2=1.197+7

x2=8.197

x=±2.86

· Solve the equation: 73x+47x-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): 73x+4+x-3=8

74x+1=8

Step 1: Take logs of both sides: log74x+1=log8

Step 2: Bring the power to the front: 4x+1log7=log8

Step 3: Tidy up and solve as normal: 4x+1=log8log7

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)+470.01715-3=8 ü

(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: log53x-4y=log25

Step 2: Bring the power to the front: 3x-4ylog5=log25

Step 3: Tidy up: 3x-4y=2

Equation 2:

Step 1: Take logs of both sides: log6-2x+4y=log1

Step 2: Bring the power to the front: -2x+4ylog6=log1

Step 3: Tidy up: -2x+4y=0

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

Add the equations to get: x=2

Substitute in equation 1: 32-4y=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. log33 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: lognn=1

We could use this fact to solve this type of equation.

· Solve the equation: 43x+2=7

Step 1: Take the log of both sides using 4 as the base: log443x+2=log47

Step 2: use the log rule to bring the power to the front: 3x+2log44=log47

Step 3: we know that log44=1 so we now have: 3x+2(1)=log47

Step 4: Tidy up and solve the equation as normal. 3x+2=1.404

3x=1.404-2

3x=-0.596

x=-0.199