### Video Transcript

Determine the solution set of the equation 36 to the power of π₯ minus 40 times six to the power of π₯ plus 144 equals zero. Give the values correct to two decimal places.

Now, at first glance, this might look like a really complicated equation to solve. The key to solving it though is spotting that we can write 36 as six squared. And this means, in turn, we can write 36 to the power of π₯ as six squared to the power of π₯. Well, one of our laws of exponents says that we can multiply the exponents in this case. So we get six to the power of two π₯. And now, weβre going to do something a little bit strange. Weβre going to write this as six to the power of π₯ all squared. And now we can rewrite our entire equation. We write it as six π₯ all squared minus 40 times six to the power of π₯ plus 144 equals zero. And now we might spot that we actually have a special type of quadratic equation.

Weβll perform a substitution to make this a little more obvious. Weβre going to let π¦ be equal to six to the power of π₯. So our equation becomes π¦ squared minus 40π¦ plus 144 equals zero. And to solve a quadratic equation, we know one of the things we can do is factor the quadratic expression. We know that a quadratic expression can be factored into the product of two binomials. The first term in each binomial must be equal to π¦ since π¦ times π¦ gives us the π¦ squared weβre looking for. To find the other term, we need to find two numbers whose product is 144 and whose sum is negative 40. Well, the only two numbers that satisfy this criteria are negative 36 and negative four. And so we find that π¦ minus 36 times π¦ minus four is equal to zero.

But, of course, π¦ minus 36 and π¦ minus four are just simply numbers. And these numbers have a product of zero. So what can we say about the numbers π¦ minus 36 and π¦ minus four? Well, either π¦ minus 36 must be equal to zero or π¦ minus four must be equal to zero. And if we solve each of these equations for π¦, we find that π¦ is equal to 36 or π¦ is equal to four. Weβre not quite finished though. We wanted to determine the solution set of an equation in π₯. So weβre going to go back to our earlier substitution. We said that π¦ was equal to six to the power of π₯. And so we created two further equations in π₯, either six to the power of π₯ equals 36 or six to the power of π₯ equals four.

Well, we know that the power of six that gives us 36 is two. But what about the power of six that gives us four? Thatβs not instantly obvious. So what weβre going to do is take logs of both sides of this equation. So we find that log of six to the power of π₯ is equal to log of four. Now, technically, when we donβt write the logs, weβre assuming that itβs log base 10. But it really doesnβt matter as long as they are the same. And then weβre going to use one of our laws of logs. And this says that log of π to the power of π is the same as π times the log of π. So π₯ times log of six equals log of four. And we can solve this equation for π₯ by dividing through by log of six. So π₯ is equal to log of four over log of six, which is 0.7737 and so on.

Remember, if we type into our calculator, we might need to type log base 10 of four over log base 10 of six. But we could equally have chosen log base two of four over log base two of six as long as the bases were the same. And so we round each of our solutions to two decimal places. And we represent the set of these solutions using the curly brackets. And, of course, we could check these by substituting them back into our original equation, remembering that since we rounded the second solution 0.77, we might get a value thatβs ever so slightly not zero.

The solution set of the equation 36 to the power of π₯ minus 40 times six to the π₯ power plus 144 equals zero is 2.00 and 0.77.