4.06 Equations with logarithms

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

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​(x−4)=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​(x−4)	$=$=	$3^2$32	Using inverse operations and raising both sides into the exponent of a base $3$3
$x-4$x−4	$=$=	$9$9	Simplify using the inverse property $a^{\log_ax}=x$aloga​x=x
$x$x	$=$=	$13$13	Solving

 

example 2

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

Think: We will first rearrange our equation into the form $\log_af(x)=c$loga​f(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​(2x−3)−8	$=$=	$0$0
$4\log_5(2x-3)$4log5​(2x−3)	$=$=	$8$8	Rearranging
$\log_5(2x-3)$log5​(2x−3)	$=$=	$2$2	Rearranging and simplifying
$2x-3$2x−3	$=$=	$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)$2logx−log(x−3)=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  $\log_am=\log_an$loga​m=loga​n if and only if $m=n$m=n.

Do:

$\log x^2-\log\left(x-3\right)$logx2−log(x−3)	$=$=	$\log\left(x+4\right)$log(x+4)	Using the power property of logarithms to simplify
$\log\left(\frac{x^2}{x-3}\right)$log(x2x−3​)	$=$=	$\log(x+4)$log(x+4)	Using the division property of logarithms to simplify
$\frac{x^2}{x-3}$x2x−3​	$=$=	$x+4$x+4	Equating equivalent parts
$x^2$x2	$=$=	$\left(x-3\right)\left(x+4\right)$(x−3)(x+4)	Rearranging
$x^2$x2	$=$=	$x^2+x-12$x2+x−12	Expanding and simplifying
$x$x	$=$=	$12$12	Solving

 

 

Practice questions

question 1

question 2

 

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 $\log_ba$logb​a) 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

 

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	$=$=	$\log7-5\log3$log7−5log3	Rearranging
$x$x	$=$=	$\frac{\log7-5\log3}{2\log3}$log7−5log32log3​	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)	$=$=	$\log_37$log3​7	Taking log base $3$3 of both sides
$2x+5$2x+5	$=$=	$\log_37$log3​7	Simplify using the inverse property $\log_aa^x=x$loga​ax=x
$2x$2x	$=$=	$\log_37-5$log3​7−5	Rearranging
$x$x	$=$=	$\frac{\log_37-5}{2}$log3​7−52​	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}$24x−1=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(24x−1)	$=$=	$\log\left(3^{2x+2}\right)$log(32x+2)	Taking logs of both sides
$\left(4x-1\right)\log2$(4x−1)log2	$=$=	$\left(2x+2\right)\log3$(2x+2)log3	Using the power property on each side
$4x\log2-\log2$4xlog2−log2	$=$=	$2x\log3+2\log3$2xlog3+2log3	Expanding
$4x\log2-2x\log3$4xlog2−2xlog3	$=$=	$2\log3+\log2$2log3+log2	Rearranging, bringing terms involving $x$x to the left-hand side
$x\left(4\log2-2\log3\right)$x(4log2−2log3)	$=$=	$2\log3+\log2$2log3+log2	Factorising out $x$x
$x$x	$=$=	$\frac{2\log3+\log2}{4\log2-2\log3}$2log3+log24log2−2log3​	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

question 5

Question 6