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Lesson Explainer: Natural Logarithmic Equations Mathematics • 10th Grade

In this explainer, we will learn how to use natural logarithms to solve exponential and logarithmic equations.

Recall that a logarithm is simply the power a number must be raised to in order to give a certain value. For example, the logarithmic equation log๏Œบ๐‘=๐‘ is another way of writing the exponential equation ๐‘Ž=๐‘๏Œผ. We refer to ๐‘Ž as the base, ๐‘ as the power that ๐‘Ž is raised to (the exponent), and ๐‘ as the answer when we raise ๐‘Ž to the power of ๐‘. To illustrate this principle, the logarithmic equation log๏Šง๏Šฆ1000=3 is another way of writing the exponential equation 10=1000๏Šฉ.

We should already be familiar with the properties of logarithms (often called โ€œlogsโ€), including the laws of logarithms, which help us to simplify expressions containing logs. We recap these laws here, as they will prove extremely useful when solving exponential and logarithmic equations.

Laws: Logarithms

For real numbers ๐‘ฅ>0, ๐‘ฆ>0, ๐‘Ž>0, and ๐‘˜, with ๐‘Žโ‰ 1, the following laws of logarithms hold: MultiplicationlawlogloglogDivisionlawlogloglogPowerlawloglogโˆถ๐‘ฅ+๐‘ฆ=(๐‘ฅ๐‘ฆ)โˆถ๐‘ฅโˆ’๐‘ฆ=๏€ฝ๐‘ฅ๐‘ฆ๏‰โˆถ๏€น๐‘ฅ๏…=๐‘˜๐‘ฅ๏Œบ๏Œบ๏Œบ๏Œบ๏Œบ๏Œบ๏Œบ๏‡๏Œบ

We will also need to know the following special cases: IdentitylawlogforZerolawlogforPowerlawwhenlogloglogโˆถ๐‘Ž=1,๐‘Ž>0,๐‘Žโ‰ 1โˆถ1=0,๐‘Ž>0,๐‘Žโ‰ 1๐‘˜=โˆ’1โˆถ๏€ผ1๐‘ฅ๏ˆ=๏€น๐‘ฅ๏…=โˆ’๐‘ฅ๏Œบ๏Œบ๏Œบ๏Œบ๏Šฑ๏Šง๏Œบ

Now, recall that the inverse of a function does the opposite to the function; for a function ๐‘“(๐‘ฅ), we use the notation ๐‘“(๐‘ฅ)๏Šฑ๏Šง to represent its inverse. We know that the inverse of any exponential function ๐‘“(๐‘ฅ)=๐‘Ž๏— is given by the logarithmic function ๐‘“(๐‘ฅ)=๐‘ฅ๏Šฑ๏Šง๏Œบlog. Therefore, the advantage of a logarithm is that it โ€œundoesโ€ exponentiation, which allows us to solve exponential equations.

In the specific case where the base is ๐‘’, we have the natural exponential functionย ๐‘“(๐‘ฅ)=๐‘’๏—, with its inverse given by ๐‘“(๐‘ฅ)=๐‘ฅ๏Šฑ๏Šง๏Œพlog. We refer to log๏Œพ๐‘ฅ as the natural logarithmic function and usually write it with the notation ln๐‘ฅ.

For any function with an inverse, the graph of the inverse function is a reflection of the original function in the line ๐‘ฆ=๐‘ฅ. Therefore, the graph of ๐‘ฆ=๐‘ฅln is a reflection of the graph of ๐‘ฆ=๐‘’๏— in the line ๐‘ฆ=๐‘ฅ, as shown below.

The graph of ๐‘ฆ=๐‘ฅln also has the following specific properties:

  • It crosses the ๐‘ฅ-axis at (1,0), which tells us that ln1=0.
  • The ๐‘ฆ-axis is an asymptote, which shows that ๐‘ฆ=๐‘ฅln is defined only for positive values of ๐‘ฅ and does not exist for ๐‘ฅโ‰ค0.
  • As ๐‘ฅโ†’โˆž, then ln๐‘ฅโ†’โˆž too, though ln๐‘ฅ grows very slowly.

(Note that, for different values of ๐‘Ž, the graphs of ๐‘ฆ=๐‘Ž๏— and ๐‘ฆ=๐‘ฅlog๏Œบ would look very similar to this, but with steeper or shallower curves, depending on the value of ๐‘Ž.)

Since we know that applying the inverse of a function โ€œundoesโ€ the action of the original function, then writing ๐‘“(๐‘ฅ)=๐‘’๏— and ๐‘“(๐‘ฅ)=๐‘ฅ๏Šฑ๏Šงln, we have ๐‘“(๐‘“(๐‘ฅ))=๐‘“(๐‘’)=(๐‘’)=๐‘ฅ,๐‘“๏€น๐‘“(๐‘ฅ)๏…=๐‘“(๐‘ฅ)=๐‘’=๐‘ฅ.๏Šฑ๏Šง๏Šฑ๏Šง๏—๏—๏Šฑ๏Šง๏—lnlnln

We will use the two key facts that ln(๐‘’)=๐‘ฅ๏— and ๐‘’=๐‘ฅln๏— throughout this explainer.

Having built up the necessary concepts, we can now tackle some equations. For instance, suppose we are asked to solve the exponential equation ๐‘’=9๏Šจ๏—, giving our answer to three decimal places.

To solve for ๐‘ฅ, we need a means of extracting 2๐‘ฅ from its position as the power (or exponent) of ๐‘’. Now, we already know that we can solve exponential equations such as 2=8๏— by taking the logarithm to base 2 of both sides. We would get loglog๏Šจ๏—๏Šจ2=8, which simplifies to ๐‘ฅ=3. Therefore, since taking the natural logarithm is the inverse operation of raising ๐‘’ to a power, here our first step is to take natural logarithms (often called โ€œtaking logsโ€) of both sides of the equation. Then, we simplify the result.

Taking logs and then dividing both sides by 2 gives lnlnlnln๏€น๐‘’๏…=92๐‘ฅ=9๐‘ฅ=129.๏Šจ๏—

Using our calculator, we get ๐‘ฅ=1.098612โ€ฆ. Finally, by rounding to three decimal places, we obtain the answer ๐‘ฅ=1.099.

Note that if we had been asked to give our answer in exact form, we would have left the value of ๐‘ฅ as the exact logarithmic expression 129ln. It is important to pay careful attention to the wording of questions to make sure that we always give our answer in its required form.

Let us now try some examples to practice these skills.

Example 1: Solving Natural Exponential Equations Using Natural Logarithms

Find, to the nearest thousandth, the value of ๐‘ฅ such that ๐‘’=19๏Šช๏—๏Šฑ๏Šฉ.

Answer

Recall that the natural logarithmic function ๐‘ฆ=๐‘ฅlog๏Œพ, usually written as ๐‘ฆ=๐‘ฅln, is the inverse of the natural exponential function ๐‘ฆ=๐‘’๏—. We can use the fact that ln(๐‘’)=๐‘ฅ๏—, together with the laws of logarithms, to help us solve natural exponential equations.

Here, we have the natural exponential equation ๐‘’=19๏Šช๏—๏Šฑ๏Šฉ. Since taking the natural logarithm is the inverse operation of raising ๐‘’ to a power, our first step is to take natural logarithms (often called โ€œtaking logsโ€) of both sides of the equation: lnlnln๏€น๐‘’๏…=194๐‘ฅโˆ’3=19.๏Šช๏—๏Šฑ๏Šฉ

We solve for ๐‘ฅ by adding 3 to both sides and then dividing both sides by 4, which gives 4๐‘ฅ=19+3๐‘ฅ=14(19+3).lnln

Using our calculator, we get ๐‘ฅ=1.486109โ€ฆ. We were asked to round our answer to the nearest thousandth. Remember that the thousandths digit is the third digit after the decimal point, which, in this case, is a 6. The digit following this (the ten-thousandths digit) is a 1, so the answer rounds down to ๐‘ฅ=1.486 to the nearest thousandth.

We may be asked to solve more complex natural exponential equations. For example, suppose we have an equation such as ๐‘’+3๐‘’โˆ’10=0.๏Šจ๏—๏—

With questions like this, the trick is to realize that ๐‘’๏Šจ๏— is the same as (๐‘’)๏—๏Šจ. This means that the left-hand side of the given equation is actually a quadratic function of ๐‘’๏—. Hence, to solve the equation, we start by using the substitution ๐‘ง=๐‘’๏— to get ๐‘ง+3๐‘งโˆ’10=0.๏Šจ

As this is now a quadratic equation in ๐‘ง, we will factor the equation and solve for ๐‘ง. Then, we can work backward from the fact that ๐‘ง=๐‘’๏— to find the corresponding value (or values) of ๐‘ฅ.

To factor ๐‘ง+3๐‘งโˆ’10๏Šจ, we need to identify factor pairs that multiply to give โˆ’10 and then select the pair that adds to give 3. It is easy to check that the required numbers are โˆ’2 and 5, so we can factor ๐‘ง+3๐‘งโˆ’10๏Šจ to obtain (๐‘งโˆ’2)(๐‘ง+5). Thus, our equation becomes (๐‘งโˆ’2)(๐‘ง+5)=0, so ๐‘ง=2 or ๐‘ง=โˆ’5. That is, ๐‘’=2๏— or ๐‘’=โˆ’5๏—.

Now, remember that ๐‘’>0๏— for all real values of ๐‘ฅ, which can be verified by referring back to our diagram showing the relationship between ๐‘ฆ=๐‘’๏— and ๐‘ฆ=๐‘ฅln. Therefore, the equation ๐‘’=โˆ’5๏— has no solution, so we are left to solve ๐‘’=2๏—. This is now in the form of a simple natural exponential equation of the sort that we solved earlier. Taking logs of both sides gives the exact solution ๐‘ฅ=2ln.

Let us test our understanding by tackling a similar type of example.

Example 2: Solving Natural Exponential Equations Using Natural Logarithms

If ๐‘’โˆ’2๐‘’โˆ’3=0๏Šจ๏—๏—, where ๐‘ฅ is a real number, find all possible values of ๐‘’๏—.

Answer

Recall that the natural logarithmic function ๐‘ฆ=๐‘ฅlog๏Œพ, usually written as ๐‘ฆ=๐‘ฅln, is the inverse of the natural exponential function ๐‘ฆ=๐‘’๏—. We can use the fact that ln(๐‘’)=๐‘ฅ๏—, together with the laws of logarithms, to help us solve natural exponential equations.

In this question, we have the natural exponential equation ๐‘’โˆ’2๐‘’โˆ’3=0๏Šจ๏—๏—, and we need to find all possible values of ๐‘’๏—. Notice that the left-hand side is a quadratic function of ๐‘’๏—, so we start by using the substitution ๐‘ง=๐‘’๏— to get ๐‘งโˆ’2๐‘งโˆ’3=0.๏Šจ

As this is now a quadratic equation in ๐‘ง, we will factor the equation and solve for ๐‘ง.

To factor ๐‘งโˆ’2๐‘งโˆ’3๏Šจ, we need to identify factor pairs that multiply to give โˆ’3 and then select the pair that adds to give โˆ’2. It is easy to check that the required numbers are โˆ’3 and 1, so we can factor ๐‘งโˆ’2๐‘งโˆ’3๏Šจ to obtain (๐‘งโˆ’3)(๐‘ง+1). Thus, our equation becomes (๐‘งโˆ’3)(๐‘ง+1)=0, so ๐‘ง=3 or ๐‘ง=โˆ’1. That is, ๐‘’=3๏— or ๐‘’=โˆ’1๏—.

Now, remember that ๐‘’>0๏— for all real values of ๐‘ฅ. Therefore, we cannot have ๐‘’=โˆ’1๏—, so we are left with the single solution ๐‘’=3๏—.

Next, suppose that we are asked to solve the natural logarithmic equation ln(3๐‘ฅโˆ’1)=2, giving our answer in exact form.

This time, to solve for ๐‘ฅ, we need a means of extracting 3๐‘ฅโˆ’1 from its position within a natural logarithm. Since raising ๐‘’ to a power is the inverse operation of taking the natural logarithm, our first step is to raise ๐‘’ to the power of both sides of the equation. Then, we simplify the result by adding 1 to both sides, followed by dividing through by 3: ๐‘’=๐‘’3๐‘ฅโˆ’1=๐‘’3๐‘ฅ=๐‘’+1๐‘ฅ=13๏€น๐‘’+1๏….ln(๏Šฉ๏—๏Šฑ๏Šง)๏Šจ๏Šจ๏Šจ๏Šจ

Our next example features a natural logarithmic equation.

Example 3: Solving Equations Involving Natural Logarithms

Solve lnln9=74๏Šญ๏—๏Šฑ๏Šฏ for ๐‘ฅ, giving your answer to the nearest hundredth.

Answer

Recall that the natural exponential function ๐‘ฆ=๐‘’๏— is the inverse of the natural logarithmic function ๐‘ฆ=๐‘ฅlog๏Œพ, which is usually written as ๐‘ฆ=๐‘ฅln. We can use the fact that ๐‘’=๐‘ฅln๏—, together with the laws of logarithms, to help us solve natural logarithmic equations.

To solve the natural logarithmic equation lnln9=74๏Šญ๏—๏Šฑ๏Šฏ for ๐‘ฅ, we first need a means of extracting 7๐‘ฅโˆ’9 from its position as the power of 9 within the natural logarithm on the left-hand side of the equation. If we apply the power law of logarithms loglog๏Œบ๏‡๏Œบ๏€น๐‘ง๏…=๐‘˜๐‘ง with ๐‘Ž=๐‘’, ๐‘ง=9, and ๐‘˜=7๐‘ฅโˆ’9, we find that lnln9=(7๐‘ฅโˆ’9)9๏Šญ๏—๏Šฑ๏Šฏ. Therefore, our equation becomes (7๐‘ฅโˆ’9)9=74.lnln

Next, we divide both sides by ln9 to get 7๐‘ฅโˆ’9=749.lnln

Finally, we add 9 to both sides, followed by dividing through by 7: 7๐‘ฅ=749+9๐‘ฅ=17๏€ฝ749+9๏‰.lnlnlnln

Using our calculator, we get ๐‘ฅ=1.565552โ€ฆ. We were asked to round our answer to the nearest hundredth. Remember that the hundredths digit is the second digit after the decimal point, which in this case is a 6. The digit following this (the thousandths digit) is a 5, so the answer rounds up to ๐‘ฅ=1.57 to the nearest hundredth.

Sometimes, we meet exponential equations that include a combination of exponential expressions. In cases like this, we might need to apply several different operations before we get to the desired solution. Here is an example of this type.

Example 4: Solving Equations Using Natural Logarithms

Solve the equation 2๐‘’=7๏—๏Šฉ๏—๏Šฑ๏Šง, giving your answer in the form ๐‘Ž+๐‘๐‘+๐‘‘lnln.

Answer

Recall that the natural logarithmic function ๐‘ฆ=๐‘ฅlog๏Œพ, usually written as ๐‘ฆ=๐‘ฅln, is the inverse of the natural exponential function ๐‘ฆ=๐‘’๏—. We can use the fact that ln(๐‘’)=๐‘ฅ๏—, together with the laws of logarithms, to help us solve exponential equations.

In this question, we have the exponential equation 2๐‘’=7๏—๏Šฉ๏—๏Šฑ๏Šง, where the left-hand side is the product of the exponential function 2๏— and the natural exponential function ๐‘’๏Šฉ๏—๏Šฑ๏Šง. We know that taking the natural logarithm is the inverse operation of raising ๐‘’ to a power. Even though only one of the left-hand exponential functions has base ๐‘’, it is still a sensible first step to take logs of both sides of the equation. Therefore, we get lnln๏€น2๐‘’๏…=7.๏—๏Šฉ๏—๏Šฑ๏Šง

Next, since the expression on the left-hand side is now the log of a product of two exponential functions, we can apply the multiplication law of logarithms logloglog๏Œบ๏Œบ๏Œบ๐‘ค+๐‘ง=(๐‘ค๐‘ง) in reverse, with ๐‘Ž=๐‘’, ๐‘ค=2๏—, and ๐‘ง=๐‘’๏Šฉ๏—๏Šฑ๏Šง. This enables us to separate the two exponential functions and then simplify the second one, as follows: lnlnlnlnln(2)+๏€น๐‘’๏…=7(2)+3๐‘ฅโˆ’1=7.๏—๏Šฉ๏—๏Šฑ๏Šง๏—

Note that we can then simplify the first term on the left-hand side by applying the power law of logarithms loglog๏Œบ๏‡๏Œบ๏€น๐‘ง๏…=๐‘˜๐‘ง with ๐‘Ž=๐‘’, ๐‘ง=2, and ๐‘˜=๐‘ฅ. Thus, our equation becomes ๐‘ฅ2+3๐‘ฅโˆ’1=7.lnln

To solve this equation for ๐‘ฅ, first, we add 1 to both sides to collect the terms involving ๐‘ฅ on one side and the number terms on the opposite side: ๐‘ฅ2+3๐‘ฅ=7+1.lnln

Then, we factor the left-hand side to get ๐‘ฅ(2+3)=7+1.lnln

Lastly, we divide through by ln2+3, which gives ๐‘ฅ=7+12+3=1+73+2.lnlnlnln

As we now have ๐‘ฅ in the form ๐‘Ž+๐‘๐‘+๐‘‘lnln, this is our answer.

Our final example involves a natural exponential equation taken from a real-life context.

Example 5: Solving Real-Life Problems Using Natural Logarithms

The number of people infected with measles during an outbreak in a country can be modeled by the exponential equation ๐‘(๐‘ก)=150๐‘’๏Šฆ๏Ž–๏Šฆ๏Šฌ๏, where ๐‘(๐‘ก) is the number of people infected after ๐‘ก days.

  1. How many people were infected at the start of the outbreak?
  2. How many people are infected after 30 days, rounding to the nearest person?
  3. How many days does it take for the number of people infected to double?

Answer

Recall that the natural logarithmic function ๐‘ฆ=๐‘ฅlog๏Œพ, usually written as ๐‘ฆ=๐‘ฅln, is the inverse of the natural exponential function ๐‘ฆ=๐‘’๏—. We can use the fact that ln(๐‘’)=๐‘ฅ๏—, together with the laws of logarithms, to help us solve natural exponential equations.

Here, we are given the natural exponential equation ๐‘(๐‘ก)=150๐‘’๏Šฆ๏Ž–๏Šฆ๏Šฌ๏, which gives the number of people infected with measles ๐‘ก days after the start of an outbreak in a particular country.

Part 1

To work out the number of people infected at the start of the outbreak, we substitute ๐‘ก=0 into the equation, which gives ๐‘(0)=150๐‘’=150๐‘’=150.๏Šฆ๏Ž–๏Šฆ๏Šฌร—๏Šฆ๏Šฆ

Part 2

To work out the number of people infected 30 days after the start of the outbreak, we substitute ๐‘ก=30 into the equation, which gives ๐‘(30)=150๐‘’=150๐‘’=907.447โ€ฆ.๏Šฆ๏Ž–๏Šฆ๏Šฌร—๏Šฉ๏Šฆ๏Šง๏Ž–๏Šฎ

As our answer represents a number of people, we must round it to the nearest whole number. Since the first digit after the decimal point is a 4, we round down to the nearest integer, which gives an answer of 907.

Part 3

To work out how many days it takes for the number of people infected to double, observe that this is equivalent to finding the value of ๐‘ก for which ๐‘(๐‘ก) is double the number infected at the start, (i.e., ๐‘(๐‘ก)=2ร—150=300). Therefore, we must solve the natural exponential equation 300=150๐‘’.๏Šฆ๏Ž–๏Šฆ๏Šฌ๏

First, we divide both sides by 150 to get 2=๐‘’.๏Šฆ๏Ž–๏Šฆ๏Šฌ๏

Next, since taking the natural logarithm is the inverse operation of raising ๐‘’ to a power, we take logs of both sides of the equation, which gives lnlnln2=๏€น๐‘’๏…2=0.06๐‘ก.๏Šฆ๏Ž–๏Šฆ๏Šฌ๏

Then, dividing through by 0.06, we have lnln20.06=๐‘ก๐‘ก=20.06.

Using our calculator, we get ๐‘ก=11.552โ€ฆ. As our answer represents a number of days, it must be a whole number. Since 11.552โ€ฆ>11, this tells us that it takes more than 11 days for the number of people infected to double. Hence, we round up to the nearest integer, so the required number of days is 12.

Let us finish by recapping some key concepts from this explainer.

Key Points

  • The natural logarithmic function is ๐‘ฆ=๐‘ฅlog๏Œพ, which is usually written as ๐‘ฆ=๐‘ฅln. It is the inverse of the natural exponential function ๐‘ฆ=๐‘’๏—.
  • We can use the fact that ln(๐‘’)=๐‘ฅ๏— and ๐‘’=๐‘ฅln๏—, together with the laws of logarithms, to solve exponential and logarithmic equations.

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