Using the properties we have explore so far, we can now solve a variety of equations using and involving logarithms.
It's important to recall that exponentiation and logarithms are the inverse of each other and have the following properties:
From previous properties we have explored, we have:
$\log_aa^x=x$logaax=x
And 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, or 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: The inverse operation of a logarithm, is exponentiation, and in this case, we can raise 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 $a^{\log_ax}=x$alogax=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 $\log_af(x)=c$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(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 as a single logarithm. We can then use the inverse operations and raise both sides into the exponent of $10$10 or more simply we can equate expressions within the logarithm on each side and solve. Since, $\log_am=\log_an$logam=logan 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
Solve $\log_6\left(3x-9\right)=2$log6(3x−9)=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(x−3)=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 | $=$= | $\log7-5\log3$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) | $=$= | $\log_37$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 | $=$= | $\log_37-5$log37−5 |
Rearranging |
| $x$x | $=$= | $\frac{\log_37-5}{2}$log37−52 |
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
Consider the equation $2^{9x-4}=90$29x−4=90.
Make $x$x the subject of the equation.
Evaluate $x$x to three decimal places.
question 5
Solve $2^{3x}=3^{7x-1}$23x=37x−1, 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.