# Question Video: Solving Exponential Equations by Factorisation Mathematics

Find the solution set of 3^(2𝑥) + 243 = 244 × 3^(𝑥) in ℝ.

02:18

### Video Transcript

Find the solution set of three to the power of two 𝑥 plus 243 equals 244 times three to the power of 𝑥 in the set of real numbers.

The solution set is the set of all values of 𝑥 that satisfy this equation. Now, by spotting that three to the power of two 𝑥 is equal to three to the power of 𝑥 all squared, we can form a quadratic equation. We perform a substitution. We let 𝑦 be equal to three to the power of 𝑥. And our equation becomes 𝑦 squared plus 243 equals 244𝑦.

Now, we know that, to solve a quadratic equation, we need to set it equal to zero. So, we’re going to subtract 244𝑦 from both sides. And so, we have the equation 𝑦 squared minus 244𝑦 plus 243 equals zero. This is quite a nice equation to solve because we can simply factor the left-hand side.

We know that the first term in each binomial must be 𝑦 since 𝑦 times 𝑦 is 𝑦 squared. Then, we’re looking for two numbers who have a product of 243 and whose sum is the coefficient of 𝑦. So, that’s negative 244. Well, those two numbers are negative one and negative 243. So, our equation becomes 𝑦 minus one times 𝑦 minus 243 equal zero.

Now, the product of these two binomials is zero. So, for this to be the case, either 𝑦 minus one must itself be equal to zero or 𝑦 minus 243 must be equal to zero. We’ll solve each of these equations for 𝑦. We solve the first equation by adding one to both sides. And we get 𝑦 equals one. Similarly, for our second equation, we add 243. So, 𝑦 is equal to 243.

But remember, we were looking to solve for 𝑥. So, we go back to our substitution. So, in each case, let’s replace 𝑦 with three to the power of 𝑥. So, we get three to the power of 𝑥 equals one and three to the power of 𝑥 equals 243. Well, we know that anything to the power of zero is one. So, in this first equation, 𝑥 must be equal to zero. And we know three to the fifth power is 243. So, we get 𝑥 equals five as our second solution.

And so, there are two values in the solution set to our equation. They are zero and five.