Video: Solving Exponential Equations Using Logarithms

In this video, we will learn how to use logarithms to solve exponential equations.

17:27

Video Transcript

In this video, we will learn how to use logarithms to solve exponential equations. We will begin by looking at the link between exponential functions and logarithmic functions. We will also recall the laws of logarithms.

We know that logarithmic functions are the inverses of exponential functions. If π‘Ž to the power of π‘₯ is equal to 𝑏, then π‘₯ is equal to log base π‘Ž of 𝑏. We will use this rule to solve equations in exponential form. For example, if two to the power of π‘₯ is equal to 16, π‘₯ is equal to log base two of 16. Using our scientific calculator, this gives us an answer of four. We know this is correct as two to the power of four is 16.

In this video, when solving exponential equations, we will also need to consider the laws of logarithms. The first law states that log base π‘Ž of π‘₯ plus log base π‘Ž of 𝑦 is equal to log base π‘Ž of π‘₯ multiplied by 𝑦. In the same way, log base π‘Ž of π‘₯ minus log base π‘Ž of 𝑦 is equal to log base π‘Ž of π‘₯ divided by 𝑦. Finally, we have log base π‘Ž of π‘₯ to the power of 𝑛 is equal to 𝑛 multiplied by log base π‘Ž of π‘₯.

It is this third law together with the link between exponential equations and logarithmic equations that we will use for the majority of this video. We notice that in all three laws of logarithms, the base must be the same. We recall that log base 10 of π‘₯ is usually just written as log of π‘₯. It is therefore common that when taking logarithms, we take logs to the base 10. This means that we do not need to write the base in every line of our calculations. We will now look at some specific questions where we need to solve exponential equations.

Solve three to the power of π‘₯ equals 11 for π‘₯, giving your answer to three decimal places.

There’re two ways that we can solve this equation using logarithms. Firstly, we can use the fact that if π‘Ž to the power of π‘₯ is equal to 𝑏, then π‘₯ is equal to log base π‘Ž of 𝑏. In this question, the constants π‘Ž and 𝑏 are three and 11, respectively. This means that π‘₯ is equal to log base three of 11. We can type the right-hand side directly into our scientific calculator, giving us 2.182658 and so on. As we want the answer given to three decimal places, the deciding number is the six. When the deciding number is five or greater, we round up. Therefore, π‘₯ is equal to 2.183. We can check this answer by substituting our value back into the original equation. Three to the power of π‘₯ is equal to 11.

An alternative method to solve this question would be to take logs of both sides first. We recall that a logarithm written without a base is log base 10. One of our laws of logarithms states that log π‘₯ to the power of 𝑛 is equal to 𝑛 multiplied by log π‘₯. As the exponent on the left-hand side of our equation is π‘₯, this can be rewritten as π‘₯ multiplied by log three. This is equal to log 11. We can then divide both sides of our equation by log three such that π‘₯ is equal to log 11 divided by log three. Once again, we get an answer rounded to three decimal places of 2.183.

In our next question, the exponent will be more complicated.

Find, to the nearest hundredth, the value of π‘₯ for which two to the power of π‘₯ plus eight is equal to nine.

In this question, we want to solve an exponential equation. We will do this using our knowledge of logarithms. There are two common methods of doing this. One way is to recall that if π‘Ž to the power of π‘₯ is equal to 𝑏, then π‘₯ is equal to log base π‘Ž of 𝑏. In this question, our exponent is π‘₯ plus eight and our values of π‘Ž and 𝑏 are two and nine, respectively. The exponent π‘₯ plus eight is therefore equal to log base two of nine. Typing the right-hand side into the calculator gives us 3.169925 and so on. As we want to find the value of π‘₯, we need to subtract eight from both sides of this equation. This gives us π‘₯ is equal to negative 4.830074 and so on. We want to round to the nearest hundredth which is the same as two decimal places. π‘₯ is therefore equal to negative 4.83. We could check this answer by substituting it back into the equation two to the power of π‘₯ plus eight is equal to nine.

An alternative method would be to take logs of both sides of the original equation. We could then use the law of logarithms that states that log π‘₯ to the power of 𝑛 is equal to 𝑛 multiplied by log π‘₯. The left-hand side of the equation becomes π‘₯ plus eight multiplied by log two. This is equal to log nine. We can then divide both sides of the equation by log two such that π‘₯ plus eight is equal to log nine divided by log two. This right-hand side is actually the same as log base two of nine. If we typed it directly into the calculator, we would get 3.169925 and so on once again. We then subtract eight from both sides of this equation, giving us π‘₯ is equal to negative 4.83.

Either one of these methods is acceptable for solving an exponential equation of this type.

Our next question is more complicated as we will have an exponent on both sides of our equation.

Use a calculator to find the value of π‘₯ for which three to the power of negative four π‘₯ minus three equals eight to the power of π‘₯ plus 4.7. Give your answer correct to two decimal places.

In order to solve this exponential equation, we will begin by taking logarithms of both sides. This gives us log three to the power of negative four π‘₯ minus three is equal to log eight to the power of π‘₯ plus 4.7. We recall that one of our laws of logarithms states that log π‘₯ to the power of 𝑛 is equal to 𝑛 multiplied by log π‘₯. Bringing down the exponents on both sides of the equation, we have negative four π‘₯ minus three multiplied by log three is equal to π‘₯ plus 4.7 multiplied by log eight.

We can then distribute the parentheses or expand the brackets on both sides. The left-hand side becomes negative four π‘₯ log three minus three log three. The right-hand side becomes π‘₯ log eight plus 4.7 log eight. Two of our four terms contain π‘₯. Therefore, we need to get both of these terms on one side of the equation. We can add four π‘₯ log three and subtract 4.7 log eight from both sides of the equation such that negative three log three minus 4.7 log eight is equal to π‘₯ log eight plus four π‘₯ log three.

Our next step is to factor out π‘₯ on the right-hand side, so this becomes π‘₯ multiplied by log eight plus four log three. We can now divide both sides of the equation by log eight plus four log three so that π‘₯ is the subject. π‘₯ is equal to negative three log three minus 4.7 log eight divided by log eight plus four log three. Recalling that a log without a base means log of base 10, we can type this into our calculator such that π‘₯ is equal to negative 2.018756 and so on. Rounding this to two decimal places gives us π‘₯ is equal to negative 2.02. This is the value of π‘₯ for which three to the power of negative four π‘₯ minus three is equal to eight to the power of π‘₯ plus 4.7.

We will know look at two slightly different types of exponential equations.

Solve two multiplied by three to the power of π‘₯ is equal to five multiplied by four to the power of π‘₯ for π‘₯, giving your answer to three decimal places.

There are lots of ways of beginning this question. One way would be to divide both sides by five multiplied by three to the power of π‘₯. This means that on the left-hand side, three to the power of π‘₯ cancels and we are left with two-fifths. On the right-hand side, the fives cancel and we are left with four to the power of π‘₯ divided by three to the power of π‘₯. When the numerator and denominator of a fraction are both raised to the same power, it can be rewritten as shown. π‘Ž to the power of π‘₯ divided by 𝑏 to the power of π‘₯ is equal to π‘Ž over 𝑏 to the power of π‘₯. This means that two-fifths is equal to four-thirds to the power of π‘₯.

We could solve this equation by taking logarithms of both sides. Alternatively, we could use the fact that if π‘Ž to the power of π‘₯ is equal to 𝑏, then π‘₯ is equal to log base π‘Ž of 𝑏. Our values of π‘Ž and 𝑏, respectively, are four-thirds and two-fifths. Therefore, π‘₯ is equal to log to the base four-thirds of two-fifths. Typing this into the calculator, we get π‘₯ is equal to negative 3.185081 and so on. As we are asked to round our answer to three decimal places, π‘₯ is equal to negative 3.185.

As previously mentioned, we could have taken logarithms of both sides when we had the equation two-fifths is equal to four-thirds to the power of π‘₯. This would have given us log of two-fifths is equal to log of four-thirds to the power of π‘₯. One of our laws of logarithms states that log π‘₯ to the power of 𝑛 is equal to 𝑛 multiplied by log π‘₯. This means we could rewrite the right-hand side as π‘₯ multiplied by log of four-thirds. Dividing both sides by log of four-thirds gives us π‘₯ is equal to log of two-fifths divided by log of four-thirds. Typing this into the calculator also gives us an answer of negative 3.185. This confirms that this is the value of π‘₯ that solves the equation two multiplied by three to the power of π‘₯ is equal to five multiplied by four to the power of π‘₯.

We will now look at one final question.

Use a calculator to find the value of π‘₯ for which two to the power of π‘₯ multiplied by seven is equal to 16 multiplied by seven to the power of π‘₯ plus nine. Give your answer correct to two decimal places.

We will begin this question by taking logs of both sides of the equation. This gives us log of two to the power of π‘₯ multiplied by seven is equal to log of 16 multiplied by seven to the power of π‘₯ plus nine. We recall that one of our laws of logarithms states that log π‘₯ plus log 𝑦 is equal to log of π‘₯ multiplied by 𝑦. This means we can rewrite the left-hand side as log two to the power of π‘₯ plus log of seven. The left-hand side can be rewritten as log 16 plus log seven to the power of π‘₯ plus nine.

One of our other laws of logarithms states that log π‘₯ to the power of 𝑛 is equal to 𝑛 log π‘₯. We can rewrite the first term on the left-hand side as π‘₯ log two. The final term on the right-hand side can be rewritten as π‘₯ plus nine multiplied by log seven. We can distribute our parentheses here to get π‘₯ log seven plus nine log seven.

Our equation has therefore become π‘₯ log two plus log seven is equal to log 16 plus π‘₯ log seven plus nine log seven. Two of our five terms have an π‘₯ in them, and we need to ensure that these are on the same side of the equation. Subtracting log seven and π‘₯ log seven from both sides of the equation gives us π‘₯ log two minus π‘₯ log seven is equal to log 16 plus nine log seven minus log seven. The last two terms on the right-hand side can be simplified. Nine log seven minus log seven is equal to eight log seven.

We can then factor out the π‘₯ on the left-hand side so that we get π‘₯ multiplied by log two minus log seven. This is equal to log 16 plus eight log seven. Finally, we divide both sides by log two minus log seven. Typing this into the calculator, we get π‘₯ is equal to negative 14.6395 and so on. We are asked to round our answer to two decimal places. Therefore, π‘₯ is equal to negative 14.64. This is the value for π‘₯ for which two to the power of π‘₯ multiplied by seven is equal to 16 multiplied by seven to the power of π‘₯ plus nine.

We will now summarize the key points from this video. We found out in this video that logarithmic functions are the inverses of exponential functions. This means that we can solve an exponential equation by taking logarithms of both sides. To solve simple exponential equations, we can use the fact that π‘Ž to the power of π‘₯ is equal to 𝑏 implies that π‘₯ is equal to log base π‘Ž of 𝑏. We can also use our three laws of logarithms: log π‘₯ plus log 𝑦 is equal to log π‘₯𝑦, log π‘₯ minus log 𝑦 is equal to log π‘₯ divided by 𝑦, and log π‘₯ to the power of 𝑛 is equal to 𝑛 multiplied by log π‘₯.

We also recall that when a logarithm is written without a base, this is the standard logarithm base 10. Using all this knowledge of logarithms, we were able to solve exponential equations, including those where the exponents are rational numbers and where the exponents are binomials in one variable.

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