Question Video: Finding the Solution Set of an Exponential Equation | Nagwa Question Video: Finding the Solution Set of an Exponential Equation | Nagwa

Question Video: Finding the Solution Set of an Exponential Equation

Determine the solution set of 6^(π‘₯) βˆ’ 3^(π‘₯ βˆ’ 1) Γ— 2^(π‘₯ βˆ’ 6) = 573/16.

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Video Transcript

Determine the solution set of six to the power of π‘₯ minus three to the power of π‘₯ minus one times two to the power of π‘₯ minus six equals 573 over 16.

By determining the solution set, we’re looking for all the values of π‘₯ that satisfy our equation. Now, before we can do anything with this equation, it’s really sensible to perform some manipulation. And the first thing we’re going to recall is one of our laws of exponents. And this says that π‘₯ to the power of π‘Ž divided by π‘₯ to the power of 𝑏 is π‘₯ to the power of π‘Ž minus 𝑏. In other words, when dividing two numbers whose base is the same, we subtract their exponents. And this means three to the power of π‘₯ minus one is the same as three to the power of π‘₯ divided by three to the power of one or three to the power of π‘₯ divided by three.

Similarly, two to the power of π‘₯ minus six is the same as two to the power of π‘₯ divided by two to the power of six. And so we rewrite our equation as six to the power of π‘₯ minus three to the power of π‘₯ over three times two to the power of π‘₯ over two to the power of six equals 573 over 16. Now, in order to make this entire equation look a little bit nicer, we’re going to multiply through by three times two to the power of six, which actually is 192. And when we do, we get 192 times six to the power of π‘₯ minus three to the power of π‘₯ times two to the power of π‘₯ equals 6876.

Now actually, three to the power of π‘₯ times two to the power of π‘₯ can be written as three times two all to the power of π‘₯, which is, of course, six to the power of π‘₯. And so our equation is 192 times six to the power of π‘₯ minus six to the power of π‘₯ equals 6876. Really, we want to make π‘₯ the subject or at least for now six to the power of π‘₯. So we factor the expression on the left-hand side such that six to the power of π‘₯ times 192 minus one is 6876, or 191 times six to the power of π‘₯ equals 6876.

Next, let’s divide through by 191. That gives us six to the power of π‘₯ equals 36. Well, we know that six squared is 36, so that means π‘₯ is equal to two. And so there is one solution to our equation, and that’s two. Our solution set is two.

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