# Question Video: Solving Exponential Equations by Factorisation Mathematics

Find the solution set of 3^(2π₯) + 243 = 244 Γ 3^(π₯) in β.

02:18

### Video Transcript

Find the solution set of three to the power of two π₯ plus 243 equals 244 times three to the power of π₯ in the set of real numbers.

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

Now, we know that, to solve a quadratic equation, we need to set it equal to zero. So, weβre going to subtract 244π¦ from both sides. And so, we have the equation π¦ squared minus 244π¦ 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 π¦ since π¦ times π¦ is π¦ squared. Then, weβre looking for two numbers who have a product of 243 and whose sum is the coefficient of π¦. So, thatβs negative 244. Well, those two numbers are negative one and negative 243. So, our equation becomes π¦ minus one times π¦ minus 243 equal zero.

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

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

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