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VCE 12 Methods 2023

4.06 Equations with logarithms

Lesson

Using the properties we have explored so far, we can now solve a variety of equations using and involving logarithms.

Remember!

From previous properties we have explored, we have:

$\log_aa^x=x$logaax=x

Due to the inverse nature of exponentials and logarithms, we also have:

$a^{\log_ax}=x$alogax=x, for $x>0$x>0

Just like if we wanted to solve $x^2=9$x2=9 we would use the inverse operation of taking the square root of both sides.  Alternatively,  if we have $\sqrt{x+1}=4$x+1=4 we would use the inverse operation of squaring both sides.   When it comes to exponential and logarithmic equations, we can use inverse operations to solve them too.

 

Solving equations involving logarithms

 

Worked examples

example 1

Solve $\log_3(x-4)=2$log3(x4)=2 using inverse operations.

Think: In this case we can find the inverse of the logarithm by firstly raising both sides into the exponent of a base $3$3.

Do:

$3^{\log_3(x-4)}$3log3(x4) $=$= $3^2$32

Using inverse operations and raising both sides into the exponent of a base $3$3

$x-4$x4 $=$= $9$9

Simplify using the inverse property $a^{\log_ax}=x$alogax=x

$x$x $=$= $13$13

Solving

 

example 2

Solve $4\log_5(2x-3)-8=0$4log5(2x3)8=0

Think: We will first rearrange our equation into the form $\log_af(x)=c$logaf(x)=c and then transpose our equation into an exponential and solve. This is in effect using the same method as example $1$1 but removing the step of writing out raising both sides into the exponent.

Do:

$4\log_5(2x-3)-8$4log5(2x3)8 $=$= $0$0

 

$4\log_5(2x-3)$4log5(2x3) $=$= $8$8

Rearranging

$\log_5(2x-3)$log5(2x3) $=$= $2$2

Rearranging and simplifying

$2x-3$2x3 $=$= $5^2$52

Transposing into an exponential equation

$2x$2x $=$= $28$28

Rearranging

$x$x $=$= $14$14

Solving

 

Example 3

Solve $2\log x-\log(x-3)=\log(x+4)$2logxlog(x3)=log(x+4)

Think: We can first use log laws to simplify the left hand side into a single logarithm. We can then equate equivalent expressions as  $\log_am=\log_an$logam=logan if and only if $m=n$m=n.

Do:

$\log x^2-\log\left(x-3\right)$logx2log(x3) $=$= $\log\left(x+4\right)$log(x+4)

Using the power property of

logarithms to simplify

$\log\left(\frac{x^2}{x-3}\right)$log(x2x3) $=$= $\log(x+4)$log(x+4)

Using the division property of

logarithms to simplify

$\frac{x^2}{x-3}$x2x3 $=$= $x+4$x+4

Equating equivalent parts

$x^2$x2 $=$= $\left(x-3\right)\left(x+4\right)$(x3)(x+4)

Rearranging

$x^2$x2 $=$= $x^2+x-12$x2+x12

Expanding and simplifying

$x$x $=$= $12$12

Solving

 

 

Practice questions

question 1

Solve $\log_6\left(3x-9\right)=2$log6(3x9)=2.

question 2

Solve $\log_4\left(5x+11\right)=\log_4\left(x+5\right)-\log_43$log4(5x+11)=log4(x+5)log43 for $x$x.

 

Testing the solution for logarithmic equations

When solving equations involving logarithms ensure your answers make sense in context and give positive arguments (that is $a>0$a>0 for $\log_ba$logba) for any log expressions in the equation or solution. Remember that the base and argument of a logarithm are always positive (and the base cannot be $1$1). Therefore, we may find that, for some log equations, we may need to reject a solution because it results in a negative argument.

It may help to first determine the possible values the variable can take before solving the equation. Check that the solutions for our previous examples were valid, and then review the practice question below.

 

Practice question

Question 3

Solve the equation $\log_4\left(x+3\right)+\log_4\left(x-3\right)=2$log4(x+3)+log4(x3)=2.

 

Solving exponential equations using logarithms

When solving an exponential equation of the type $a^x=b$ax=b where $a$a and $b$b cannot be made into the same base, we can instead use logarithms to solve, since logarithms are the inverse operation of an exponential. We do this by taking the logarithm of both sides and simplifying to solve for $x$x. We can use the logarithm of any base, but it is common practice to use base $10$10.

 

Worked examples

example 5

Solve $3^{2x+5}=7$32x+5=7

Think: We can see that we cannot easily make both sides of the same base like we learnt to do last year, as $3$3 and $7$7 are prime. This is where it will be useful to take the logarithm of both sides to solve for $x$x. We will use base $10$10 for convenience, but note that we could choose any base.

Do:

$\log\left(3^{2x+5}\right)$log(32x+5) $=$= $\log7$log7

Taking log base $10$10 of both sides

$\left(2x+5\right)\log3$(2x+5)log3 $=$= $\log7$log7

Using the power property

$2x\log3+5\log3$2xlog3+5log3 $=$= $\log7$log7

Expanding

$2x\log3$2xlog3 $=$= $\log7-5\log3$log75log3

Rearranging

$x$x $=$= $\frac{\log7-5\log3}{2\log3}$log75log32log3

Solving for $x$x

 

Alternative: Instead of taking log base $10$10 in the first step we could take log base $3$3. Let's look at how this solution would develop.

$\log_3\left(3^{2x+5}\right)$log3(32x+5) $=$= $\log_37$log37

Taking log base $3$3 of both sides

$2x+5$2x+5 $=$= $\log_37$log37

Simplify using the inverse property $\log_aa^x=x$logaax=x

$2x$2x $=$= $\log_37-5$log375

Rearranging

$x$x $=$= $\frac{\log_37-5}{2}$log3752

Divide through by $2$2

 

Reflect: Can you demonstrate that the two different answers we obtained are in fact equivalent?

example 6

Solve $2^{4x-1}=3^{2x+2}$24x1=32x+2

Think: By first taking the logarithm of both sides, we will be able to use the power property of logarithms to rearrange each side of the equation, simplify and then solve.

Do:

$\log\left(2^{4x-1}\right)$log(24x1) $=$= $\log\left(3^{2x+2}\right)$log(32x+2)

Taking logs of both sides

$\left(4x-1\right)\log2$(4x1)log2 $=$= $\left(2x+2\right)\log3$(2x+2)log3

Using the power property on each side

$4x\log2-\log2$4xlog2log2 $=$= $2x\log3+2\log3$2xlog3+2log3

Expanding

$4x\log2-2x\log3$4xlog22xlog3 $=$= $2\log3+\log2$2log3+log2

Rearranging, bringing terms involving $x$x to the left-hand side

$x\left(4\log2-2\log3\right)$x(4log22log3) $=$= $2\log3+\log2$2log3+log2

Factorising out $x$x

$x$x $=$= $\frac{2\log3+\log2}{4\log2-2\log3}$2log3+log24log22log3

Solving

 

This final answer for $x$x could be further simplified using logarithm properties. You might like to confirm that we could write $x=\frac{\log18}{\log\left(\frac{16}{9}\right)}$x=log18log(169).

Practice questions

question 4

Consider the equation $2^{9x-4}=90$29x4=90.

  1. Make $x$x the subject of the equation.

  2. Evaluate $x$x to three decimal places.

question 5

Solve $2^{3x}=3^{7x-1}$23x=37x1, by taking the logarithm of both sides.

Leave your answer in exact form.

Question 6

The equation for the population at time $t$t is given by $Q=30\times8^{5t}$Q=30×85t. Make $t$t the subject of the equation.

Outcomes

U34.AoS2.4

solution of equations of the form f(x) = g(x) over a specified interval, where f and g are functions of the type specified in the ‘Functions, relations and graphs’ area of study, by graphical, numerical and algebraic methods, as applicable

U34.AoS2.11

apply algebraic, logarithmic and circular function properties to the simplification of expressions and the solution of equations

U34.AoS2.8

analytical, graphical and numerical approaches to solving equations and the nature of corresponding solutions (real, exact or approximate) and the effect of domain restrictions

U34.AoS2.9

apply a range of analytical, graphical and numerical processes (including the algorithm for Newton’s method), as appropriate, to obtain general and specific solutions (exact or approximate) to equations (including literal equations) over a given domain and be able to verify solutions to a particular equation or equations over a given domain

U34.AoS2.7

exponent laws and logarithm laws

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