# Video: Solving Simple Equations With the Unknown in an Exponent

Explaining how to use the rules of exponents and knowledge of powers of small integers to solve simple equations, where the unknown is in an exponent. For instance, 2^(𝑥) = 8^(𝑥 + 3) becomes 2^(𝑥) = (2³)^(𝑥 + 3), then 2^(𝑥) = 2^[3(𝑥 + 3)], so 𝑥 = 3(𝑥 + 3).

11:31

### Video Transcript

The are a family exponent type questions that require you to recall and apply the exponent rules. 𝑥 to the 𝑎 times 𝑥 to the 𝑏 equals 𝑥 to the 𝑎 plus 𝑏; that’s the addition rule. 𝑥 to the 𝑎 divided by 𝑥 to the 𝑏 equals 𝑥 to the 𝑎 minus 𝑏; that’s the subtraction rule. And 𝑥 to the power of 𝑎 all to the power of 𝑏 equals 𝑥 to the power of 𝑎 times 𝑏; that’s the multiplication rule.

Now we’ve got other videos that talk about those rules in more detail if you’re not sure about them. But in this video, we’re gonna use these rules combined with our knowledge of various powers of two, three, five, and ten to solve some problems.

So the first question is find the value of 𝑥 if two to the power of 𝑥 equals eight to the power of 𝑥 plus three. Well, at first sight this looks tricky. But hopefully we can recall that two times two times two, two cubed, is equal to eight. So I can re-express eight as a power of two. So rewriting that equation with two cubed instead of eight, we’ve got two to the power of 𝑥 is equal to two cubed all to the power of 𝑥 plus three.

Now we’ve got something to the power of something to the power of something else, so we can use the multiplication rule. So two to the power of three all to the power of 𝑥 plus three just means two to the power of three times 𝑥 plus three, in parentheses. So now I’ve got two to the power of 𝑥 is equal to two to the power of three lots of 𝑥 plus three, so those two things must be equal. Two to the power of something equals two to the power of something; those two somethings must be equal.

And now I can use the distributive law to multiply out the parentheses so that I know that 𝑥 is equal to three 𝑥 plus nine. Now if I take away 𝑥 from both sides of my equation, I’m gonna have all the 𝑥s on one side. So on the left-hand side, 𝑥 take away 𝑥 is nothing. And on the right-hand side, three 𝑥 take away 𝑥 is two 𝑥, and I’ve still got my plus nine. So now I’m trying to get 𝑥 on its own; take away nine from both sides.

And on the left-hand side, zero take away nine is negative nine. And on the right hand side, I’ve got two 𝑥 plus nine minus nine so those two cancel out so I’ve just got two 𝑥. Now we still want to know what one 𝑥 is, so I need to divide both sides by two. And two divided by two is just one, so one 𝑥 is equal to minus nine over two. Or as a mixed number, 𝑥 is equal to negative four and a half.

So there were a few key tricks in solving that equation. Firstly, we had to know that eight is the same as two cubed, so you need to know your exponents of simple numbers like two, three, four, five, ten, those few. And then we needed to know the multiplication rule, so we needed to know that two to the power of three all to the power of 𝑥 plus three was the same as two to the power of three lots of or three times 𝑥 plus three.

Then we spotted the fact that it was two to the power of 𝑥 and two the same thing to the power of three times 𝑥 plus three, so those two exponents must be equal. The rest was just simple, solving linear algebra. And we came up with our final answer, 𝑥 is minus four and a half.

Next question, solve for 𝑥: nine to the power of 𝑥 plus five is equal to twenty-seven to the power of 𝑥 minus one. Now nine and twenty-seven are both multiples of three, so three times three gives us nine and three times three times three gives us twenty-seven. So three cubed is twenty-seven. So we can replace nine and twenty-seven with numbers which have the same base for their exponents, so three squared and three cubed.

So that first line is the same as three squared to the power of 𝑥 plus five is equal to three cubed to the power of 𝑥 minus one. And now we’re gonna use the multiplication rule again. So we can rewrite that as three to the power of two times 𝑥 plus five equals three to the power of three times 𝑥 minus one.

So we’ve now got the same base for our exponents, three to the power of something is equal to three to the power of something. Now if that’s the case, those somethings must be equal so two times 𝑥 plus five must be equal to three times 𝑥 minus one. And now we can use the distributive law of multiplication to multiply out the parentheses.

And we can see that two 𝑥 plus ten is equal to three 𝑥 minus five. Now I’ve got two 𝑥 on the left-hand side and three 𝑥 on the right-hand side. If I take away two 𝑥 from both sides, I’m still gonna have a positive number of 𝑥s on the right-hand side and I won’t have any 𝑥s on the left-hand side.

So on the left-hand side, I’ve got two 𝑥 plus ten take away two 𝑥 so the two 𝑥s are gonna cancel to just leave me with ten. And on the right-hand side, I’ve got three 𝑥 take away two 𝑥 is just one 𝑥 and then I’ve still got my minus three.

So now I can add three to both sides of the equation, which gives me the thirteen on the left-hand side. And on the right-hand side, I’ve got three take away three and I’ve got my 𝑥 so that’s gonna leave me with 𝑥. So my answer is 𝑥 is equal to thirteen.

So again in this question, we had to think about the bases of our exponents. And in the first case, we were able to spot that it was three squared. And in the second case, it was three cubed. So if we can then use the multiplication rule to get the same bases in each case, we can then just compare the exponents directly, and that turns it into a piece of just linear algebra.

Now for number three, find the value of 𝑥 if one over a hundred and twenty-five to the power of 𝑥 is equal to five to the power of two 𝑥 minus three. Now our first clue here is that a hundred and twenty-five is simply five cubed, so I’m gonna re-express a hundred and twenty-five as five cubed. And then I can use the multiplication rule five to the power of three to the power of 𝑥 is just five to the power of three 𝑥.

And now I’m gonna use the fact that I know about negative exponents so five to the power of negative three 𝑥 would be the same as one over five to the three 𝑥, so I’m gonna re-express that left-hand side. So instead of writing as one over five to the power of three 𝑥, I’m gonna write it as five to the power of negative three 𝑥, so we really are testing your knowledge of exponents here. And that is equal to five to the power of two 𝑥 minus three.

So we’ve reached the situation where we’ve got the same base, five in each case, and five to the power of negative three 𝑥 is equal to five to the power of two 𝑥 plus two 𝑥 minus three. So those two exponents must be equal; negative three 𝑥 must be equal to two 𝑥 minus three.

And now we’ve got our linear algebra to solve, so I’m gonna add three 𝑥 to both sides to get myself a positive number of 𝑥s on one side of the equation. And then on the left-hand side, negative three 𝑥 plus three 𝑥 is zero and on the right-hand side two 𝑥 plus three 𝑥 is five 𝑥.

And I’ve still got the negative three, so five 𝑥 minus three. So now if I add three to both sides of the equation, on the left-hand side I’ve got zero plus three, which is three, and on the right-hand side I’ve got negative three plus three. So those two cancel out which just leaves me with five 𝑥, so three is equal to five 𝑥. So now, just divide both sides of my equation by five.

And on the right-hand side, I’ve got 𝑥 times five divided by five so those two fives are gonna cancel out so I’ve just got 𝑥. And that’s equal to three-fifths. So I’m making my answer nice and clear: 𝑥 is equal to three-fifths.

So the key steps to solving that one was spotting the fact that a hundred and twenty-five is five to the power of three so we can come up with the same base exponent that we’re working from. We then use the multiplication rule, and we had to know a little bit about negative exponents to get this in the same format. Then we took our exponents that were equal, negative three 𝑥 and two 𝑥 minus three, did a bit of linear algebra, and then came up with our answer.

Now for our last example, number four, find the value of 𝑥 if a thousand to the power of two-thirds is equal to a hundred to the power of two 𝑥 plus five.

Now we’ve got a couple of basic approaches that we could go for here. We could spot the fact that a thousand is ten cubed and a hundred is ten squared so we can make those substitutions and then use our power rules and try to equate the exponents, or we could use our knowledge of exponents to say that this means the cube root of something all squared, so the cube root of a thousand is ten and then square that and we get a hundred. And then we see that we’ll have the same bases on each side, so in fact I think that’s gonna be the quicker way to go here; so that’s the way I’m gonna go.

So I’m gonna evaluate the left-hand side. So we just re-express that instead of a hundred to the power of two-thirds we know that that’s the cube root of a thousand and then the whole thing is squared. Well the cube root of a thousand is ten, so the left-hand side is ten squared, which is just a hundred.

So we’ve got a hundred is equal to a hundred to the power of two 𝑥 plus five. Well a hundred is the same as a hundred to the power of one. So now I’ve got a a hundred to the power of one is equal to a hundred to the power of two 𝑥 plus five. So I’ve got the same base for my exponents, so I can equate those exponents.

Well that just leaves me with one is equal to two 𝑥 plus five, so just subtract five from both sides. And on the right-hand side, I’ve got five take away five is nothing, so that gives me negative four is equal to two 𝑥. Now I can divide both sides by two, which means the twos are gonna cancel on the left- on the right-hand side, and negative four over two is just negative two.

So this last example was much like the previous ones. We needed to know about exponents. In this case, we needed to know about these rational exponents up here. So that means the cube root of something all squared. And we also had to remember that just a hundred without any power or without any exponent is the same as a hundred to the power of one so that we could equate those exponents, one with two 𝑥 plus five. And then the rest was just simple linear algebra to solve.

Now some questions that have the unknown in the exponents turn out to be very much more difficult to solve. But if you spot this little trick, if you can make the bases the same and then just equate the exponents, then it can make life relatively straightforward and these questions needn’t be that difficult.