Lesson Video: Logarithmic Functions | Nagwa Lesson Video: Logarithmic Functions | Nagwa

Lesson Video: Logarithmic Functions Mathematics • Second Year of Secondary School

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In this video, we will learn how to identify, write, and evaluate a logarithmic function as an inverse of the exponential function.

12:30

Video Transcript

In this video, we will learn how to identify, write, and evaluate a logarithmic function as the inverse of an exponential function. We will begin by recalling the link between exponential and logarithmic functions.

Logarithmic functions are the inverses or opposite of exponential functions. If we consider the exponential function 𝑓 of 𝑥 is equal to 𝑎 to the power of 𝑥, the inverse of 𝑓 of 𝑥 is equal to log base 𝑎 of 𝑥. This enables us to solve exponential equations by using logarithms. If 𝑦 is equal to 𝑎 to the power of 𝑥, then 𝑥 is equal to log base 𝑎 of 𝑦. We also recall that the natural logarithm written ln of 𝑥 is the inverse of 𝑒 to the power of 𝑥. Finally, when a logarithm is written without a base, it is assumed to be base 10. Log 𝑥 is the same as log base 10 of 𝑥. We will now look at some questions involving exponential and logarithmic functions.

The function 𝑓 of 𝑥 which is equal to two 𝑒 to the power of 𝑥 plus three has an inverse of the form 𝑔 of 𝑥 is equal to ln of 𝑎𝑥 plus 𝑏. What are the values of 𝑎 and 𝑏?

In order to find the inverse of any function, we begin by replacing 𝑓 of 𝑥 with 𝑦. In this case, 𝑦 is equal to two 𝑒 to the power of 𝑥 plus three. Our next step is to rearrange this equation to make 𝑥 the subject. We begin by subtracting three from both sides so that 𝑦 minus three is equal to two 𝑒 to the power of 𝑥. We can then divide both sides of our equation by two. 𝑦 minus three over two is equal to 𝑒 to the power of 𝑥. The left-hand side can be rewritten as a half 𝑦 minus three over two. We can then take the natural logarithm of both sides as we know that ln of 𝑥 is the opposite or inverse of 𝑒 to the power of 𝑥. This gives us ln of one-half 𝑦 minus three over two is equal to 𝑥.

As we have now made 𝑥 the subject of the equation, we can know swap our 𝑦- and 𝑥-variables. The inverse of 𝑓 of 𝑥 is therefore equal to ln of a half 𝑥 minus three over two. As the inverse was denoted by 𝑔 of 𝑥, this is now in the form ln of 𝑎𝑥 plus 𝑏, where 𝑎 is equal to one-half and 𝑏 is equal to negative three over two or negative three-halves. This method can be used to calculate the inverse of any function.

In our next question, we will consider the domain and range of exponential and logarithmic functions.

Consider the function 𝑓 of 𝑥 is equal to 𝑏 to the power of 𝑥, where 𝑏 is a positive real number not equal to one. What is the domain of the inverse of 𝑓 of 𝑥?

There are a few ways of approaching this problem. One way would be to recall that exponential functions and logarithmic functions are the inverse of each other. This means that if 𝑓 of 𝑥 is equal to 𝑏 to the power of 𝑥, the inverse function is equal to log base 𝑏 of 𝑥. We are asked to find the domain of this function. The domain of any function is the set of input values. We know that we can only find the logarithm of positive values. This means that the domain of the inverse function is 𝑥 is greater than zero as the only values we can substitute into the function log base 𝑏 of 𝑥 are 𝑥 greater than zero.

An alternative method here would be to consider the graphs of our functions. The graph of 𝑓 of 𝑥 is shown. It intersects the 𝑦-axis at 𝑏 and the 𝑥-axis is an asymptote. The inverse of any function is its reflection in the line 𝑦 equals 𝑥. This means that the function log base 𝑏 of 𝑥 intersects the 𝑥-axis at 𝑏 and the 𝑦-axis is an asymptote. As the domain is the set of input values, we can see from the graph that the domain of the inverse of 𝑓 of 𝑥 is all numbers greater than zero.

A final method would be to recall that the domain of 𝑓 is equal to the range of the inverse. Likewise, the range of 𝑓 of 𝑥 is equal to the domain of the inverse. The range of any function is the set of output values. We can see from the graph that the range of 𝑓 of 𝑥 is all values greater than zero. This once again proves that the domain of the inverse function is 𝑥 is greater than zero.

Our next question involves solving a logarithmic equation.

Consider the function 𝑓 of 𝑥 is equal to log base two of three 𝑥 minus one. If 𝑓 of 𝑎 is equal to three, find the value of 𝑎.

We are told that 𝑓 of 𝑎 is equal to three, so we can begin by substituting these values into the function 𝑓 of 𝑥. This gives us log base two of three 𝑎 minus one is equal to three. We recall that logarithmic functions are the inverses of exponential functions. This means that if log base 𝑎 of 𝑦 is equal to 𝑥, then 𝑎 to the power of 𝑥 is equal to 𝑦. In this question, the base 𝑎 is equal to two, the variable 𝑦 is equal to three 𝑎 minus one, and the variable 𝑥 is equal to three. Two cubed is therefore equal to three 𝑎 minus one.

We know that two cubed is equal to eight. We can then add one to both sides of this equation so that three 𝑎 is equal to nine. Dividing both sides of this equation by three gives us 𝑎 is equal to three. If the function 𝑓 of 𝑥 is equal to log base two of three 𝑥 minus one and 𝑓 of 𝑎 is equal to three, then the value of 𝑎 is three. We could check this answer on the calculator by substituting our value back in to the original function.

In our next question, we want to find the base of a logarithmic function.

Given that the graph of the function 𝑓 of 𝑥 which is equal to log base 𝑎 of 𝑥 passes through the point 1024, five, find the value of 𝑎.

We are told that our function passes through the point with 𝑥-coordinate 1024 and 𝑦-coordinate five. The function 𝑓 of 𝑥 can be rewritten as 𝑦 is equal to log base 𝑎 of 𝑥. When dealing with functions, 𝑓 of 𝑥 and 𝑦 are interchangeable. Substituting in our values of 𝑥 and 𝑦, we have five is equal to log base 𝑎 of 1024. We know that logarithmic functions and exponential functions are the inverse of each other. This means that if 𝑥 is equal to log base 𝑎 of 𝑦, then 𝑎 to the power of 𝑥 is equal to 𝑦.

We can rewrite our equation in exponential form such that 𝑎 to the power of five is equal to 1024. We can then take the fifth root of both sides of our equation to work out the value of 𝑎. The fifth root of 1024 is equal to four. We can check this by calculating four to the fifth power. This is equal to four multiplied by four multiplied by four multiplied by four multiplied by four. Four multiplied by four is equal to 16. When we multiply this by four, we get 64. 64 multiplied by four is 256. And finally, multiplying this by four gives us 1024. If the function 𝑓 of 𝑥, which is equal to log base 𝑎 of 𝑥, passes through the point 1024, five, then the base 𝑎 is equal to four.

Our final question involves solving a logarithmic equation in a real-life context.

The pH of a solution is given by the formula pH is equal to negative log of 𝑎 sub H+, where 𝑎 sub H+ is the concentration of hydrogen ions. Determine the concentration of hydrogen ions in a solution whose pH is 8.4.

When the pH is equal to 8.4, 8.4 is equal to negative log of 𝑎 sub H+. We are trying to work out this value which is the concentration of hydrogen ions. We recall that when a logarithm is written without a base, it is assumed to be base 10. Log 𝑥 is the same as log base 10 of 𝑥. We can multiply both sides of our equation by negative one such that negative 8.4 is equal to log base 10 of 𝑎 sub H+. We know that a logarithmic function is the inverse of an exponential function. If 𝑥 is equal to log base 𝑎 of 𝑦, then 𝑎 to the power of 𝑥 is equal to 𝑦. This means that 10 to the power of negative 8.4 is equal to 𝑎 sub H+. The concentration of hydrogen ions is therefore equal to 10 to the power of negative 8.4.

We will now summarize the key points from this video. We found out in this video that logarithmic functions and exponential functions are inverses of each other. This means that if 𝑥 is equal to log base 𝑎 of 𝑦, then 𝑎 to the power of 𝑥 is equal to 𝑦. This enables us to convert between exponential and logarithmic equations. We also found out that if 𝑓 of 𝑥 is equal to 𝑒 to the power of 𝑥, then the inverse of this function is equal to the natural logarithm ln of 𝑥. The domain of 𝑓 of 𝑥 is equal to the range of the inverse function. Likewise, the range of 𝑓 of 𝑥 is equal to the domain of the inverse function. This is because 𝑓 of 𝑥 and 𝑓 minus one of 𝑥 are reflections in the line 𝑦 equals 𝑥. Finally, we recalled that when a logarithm is written without a base, it is the same as base 10. Log 𝑥 is equal to log base 10 of 𝑥.

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