Question Video: Finding the Value of a Logarithm Using Substitution | Nagwa Question Video: Finding the Value of a Logarithm Using Substitution | Nagwa

Question Video: Finding the Value of a Logarithm Using Substitution Mathematics

If 𝑔(π‘₯) is the inverse of the function 𝑓(π‘₯) = 3^(2π‘₯ βˆ’ 3), find 𝑔(π‘₯).

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Video Transcript

If 𝑔 of π‘₯ is the inverse of the function 𝑓 of π‘₯ equals three to the power of two π‘₯ minus three, find 𝑔 of π‘₯.

So what we’re gonna do in this problem first of all is we’re gonna substitute 𝑦 for 𝑓 of π‘₯. So what we’re gonna have is 𝑦 equals three to the power of two π‘₯ minus three. Okay, so what’s the next step? Well, to find the inverse of a function, the method we can use is to in fact swap our π‘₯- and 𝑦-terms. And then to find out what the inverse is, we rearrange to make 𝑦 now the subject.

Well, if we want to make 𝑦 the subject, what we can use is one of our log relationships. And that is that if we have π‘Ž equals 𝑏 to the power of 𝑐, then we know that 𝑐 is gonna be equal to log to the base 𝑏 of π‘Ž. So what we have is our π‘₯ is π‘Ž, our three is our 𝑏, and our two 𝑦 minus three is our 𝑐. So what we’re gonna get is that two 𝑦 minus three is equal to log to the base three of π‘₯. And then what we do is we add three to each side of the equation, and we get two 𝑦 equals log to base three of π‘₯ plus three. And then what we can do is divide through by two. And we want to do that because, like we said, we want to make 𝑦 the subject. And at the moment, we’ve got two 𝑦 on its own on the left-hand side.

And when we do that, what we’re gonna get is 𝑦 is equal to a half log to base three of π‘₯ plus three over two. Well, what we said at the beginning was once we finished our rearranging and changed the subject, then the new 𝑦 would be the inverse of 𝑓 of π‘₯; it’s the inverse of our original function. So therefore, what we can say is that because 𝑔 of π‘₯ is the inverse of the function 𝑓 of π‘₯ equals three to the power of two π‘₯ minus three, then, therefore, 𝑔 of π‘₯ is equal to a half log to the base three of π‘₯ plus three over two.

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