Solving exponential equations using exponent properties | High School Math | Khan Academy

Let's get some practice solving some exponential equations, and we have one right over here. We have (26^{9x + 5} = 1).

So pause the video and see if you can tell me what (x) is going to be. Well, the key here is to realize that (26^0) is equal to 1. Anything to the 0th power is going to be equal to one. Zero to the zero power we can discuss some other time, but anything other than zero to the zero power is going to be one.

So we just have to say, well, (9x + 5) needs to be equal to zero. (9x + 5) needs to be equal to zero, and this is pretty straightforward to solve. Subtract five from both sides, and we get (9x = -5). Divide both sides by nine, and we are left with (x = -\frac{5}{9}).

Let's do another one of these, and let's make it a little bit more interesting. Let's say we have the exponential equation (2^{3x + 5} = 64^{x - 7}).

Once again, pause the video and see if you can tell me what (x) is going to be or what (x) needs to be to satisfy this exponential equation.

All right, so you might at first say, "Oh, maybe (3x + 5) needs to be equal to (x - 7)," but that wouldn't work because these are two different bases. You have (2^{3x + 5}) then you have (64^{x - 7}).

So the key here is to express both of these with the same base, and lucky for us, (64) is a power of two. (2^3) is eight, so it's going to be (2^3 \times 2^3); eight times eight is sixty-four, so it's (2^6) is equal to sixty-four.

You can verify that. Take six twos and multiply them together, you’re going to get (64). This is just a little bit easier for me; eight times eight, and this is the same thing as (2^6) power is (64).

And I knew it was to the sixth power because I just added the exponents because I had the same base.

All right, so I can rewrite (64). Let me rewrite the whole thing. So this is (2^{x + 5} = 2^6), and then that to the (x - 7) power.

And to simplify this a little bit, we just have to remind ourselves that if I raise something to one power and then I raise that to another power, this is the same thing as raising my base to the product of these powers (a^{b \cdot c}).

So this equation I can rewrite as (2^{3x + 5} = 2^{6 \cdot (x - 7)}). So it's going to be (6x - (6 \cdot 7) = 42).

I'll just write the whole thing in yellow: (6x - 42). I just multiplied the (6) times the entire expression (x - 7).

And so now it's interesting. I have (2^{3x + 5}) power has to be equal to (2^{6x - 42}) power, so these need to be the same exponent. So (3x + 5) needs to be equal to (6x - 42).

So there we go; it sets up a nice little linear equation for us. (3x + 5 = 6x - 42).

Let's see, we could get all of our — since, well, I'll put all my (x)'s on the right-hand side since I have more (x)'s on the right already. So let me subtract (3x) from both sides, and let me — I want to get rid of this (42) here, so let's add (42) to both sides.

And we are going to be left with (5 + 42 = 47) is equal to (3x). Now we just divide both sides by (3), and we are left with (x = \frac{47}{3}).

(x = \frac{47}{3}), and we are done.