Question Video: Finding the Solution Set of Exponential Equations Involving Square Roots | Nagwa Question Video: Finding the Solution Set of Exponential Equations Involving Square Roots | Nagwa

Question Video: Finding the Solution Set of Exponential Equations Involving Square Roots

Determine the solution set of √(9^(𝑥) − 18 × 3^(𝑥) + 81) = 72.

03:40

Video Transcript

Determine the solution set of the square root of nine to the power of 𝑥 minus 18 times three to the power of 𝑥 plus 81 equals 72.

This question is asking us to determine the solution set of an equation in 𝑥. In other words, we’re interested in finding the values of 𝑥 that make this equation true. Now this looks particularly nasty. But there is one simple thing that we can do to make it look a little bit nicer, that is, square both sides of the equation. By squaring the left-hand side of this equation, we’re essentially performing the inverse to square rooting. So we’re just left with the values inside the square root. That’s nine to the power of 𝑥 minus 18 times three to the power of 𝑥 plus 81.

Of course, since we’ve squared the left-hand side, we need to repeat that on the right-hand side. And when we square 72, we get 5184. Next, we’ll subtract this value from both sides of our equation. Then we have an equation that’s equal to zero. That is, nine to the power of 𝑥 minus 18 times three to the power of 𝑥 minus 5103 is equal to zero.

The key to solving this equation is spotting that nine can be written as three squared. And so we can write nine to the power of 𝑥 as three squared to the power of 𝑥. So our equation becomes three squared to the power of 𝑥 minus 18 times three to the power of 𝑥 minus 5103 equals zero.

Now of course we can multiply these exponents. And we can say that three squared to the power of 𝑥 is the same as three to the power of two 𝑥. Then we can reverse this process and say, well, this must be the same as three to the power of 𝑥 squared. And so we write our equation as three to the power of 𝑥 squared minus 18 times three to the power of 𝑥 minus 5103 equals zero.

And if we look carefully, we’ll see that this looks a little bit like a quadratic equation. And so we’re going to perform a substitution. We’re going to let 𝑦 be equal to three to the power of 𝑥. When we do, we find that we can write our equation as 𝑦 squared minus 18𝑦 minus 5103 equals zero.

This equation is quite easy to solve. We need to factor the expression 𝑦 squared minus 18𝑦 minus 5103. We know that the first term in each binomial must be 𝑦 since 𝑦 times 𝑦 gives us 𝑦 squared as required. Then to find the other term in each binomial, we’re looking for two numbers whose product is negative 5103 and whose sum is negative 18. These numbers are negative 81 and 63.

So our equation becomes 𝑦 minus 81 times 𝑦 plus 63 equals zero. And 𝑦 minus 81 and 𝑦 plus 63 are simply numbers. And their product is equal to zero. So for this to be the case, either 𝑦 minus 81 must be equal to zero or 𝑦 plus 63 must be equal to zero.

Let’s solve this first equation for 𝑦 by adding 81 to both sides. So we find that 𝑦 is equal to 81. And we’ll solve this second equation by subtracting 63 from both sides. So 𝑦 is equal to negative 63.

We’re not quite finished though. We were looking to solve this equation for 𝑥. So we go back to our earlier substitution 𝑦 is equal to three to the power of 𝑥. And we can replace 𝑦 in our solutions. And we see that either three to the power of 𝑥 is equal to 81 or it’s equal to negative 63. The value of 𝑥 that makes the equation three to the power of 𝑥 equals 81 true is four. But there are no values of 𝑥 such that three to the power of 𝑥 is equal to a negative value. So there are no solutions to our second equation. This means that the solution set of our equation consists of one number only. It’s four.

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