Video Transcript
Given that line segment 𝐴𝐵 is perpendicular to line segment 𝐴𝐶 and line segment 𝐴𝐷 is perpendicular to line segment 𝐵𝐶, determine the value of 𝑥. Give your answer to the nearest tenth.
If we locate 𝑥 in our diagram, it is the base of two different log functions that measure segments of the triangle. Solving for this 𝑥-value is going to require us to synthesize quite a lot of information, so we’re going to take it in a few stages. First of all, we need to think about what we know about this figure. We have a right angle at angle 𝐶𝐴𝐵 and a right angle at 𝐴𝐷𝐶. When it happens that a line begins at a vertex and intersects one of the bases of a triangle, it’s called an altitude of that triangle. And there’s a principle of inscribed similar triangles that tells us the altitude of a right triangle creates similar triangles, which means the first thing we wanna do is identify the three similar triangles in our figure.
First, we have the large exterior triangle 𝐴𝐵𝐶. If we make a sketch of triangle 𝐴𝐵𝐶, we see that the side opposite the right angle is the hypotenuse, which is the side length 𝐵𝐶. And it’s going to have a length of log base 𝑥 of 32 plus log base 𝑥 of 1024. In addition to that, we know that the side length 𝐴𝐵 measures eight times the square root of three. Now we want to think about the triangles created inside this triangle 𝐴𝐵𝐶. We have the triangle 𝐷𝐵𝐴. Again, the hypotenuse will be the side length opposite the right angle. For triangle 𝐷𝐵𝐴, the hypotenuse will be the segment 𝐴𝐵, which is eight times the square root of three. And we know the measure of segment 𝐷𝐵 is log base 𝑥 of 1024.
We have a third similar triangle, which is triangle 𝐷𝐴𝐶. The hypotenuse is the side opposite the right angle. That will be 𝐴𝐶. However, we don’t know the value of 𝐴𝐶. For the smaller triangle 𝐷𝐴𝐶, we only know the length of one segment 𝐷𝐶. And we don’t have enough information to solve for either of the other sides. Since we’re looking for the value of 𝑥 which is the base of these log functions, we need to be able to write some kind of equation. And that means we need to think about what we can say about these similar triangles. In similar triangles, corresponding sides are proportional. We know that triangle 𝐴𝐵𝐶 is similar to triangle 𝐷𝐵𝐴, which means we can say that side length 𝐴𝐵 over side length 𝐵𝐶 will be equal to side length 𝐷𝐵 over side length 𝐵𝐴.
Using this proportion, we can set up an equation that will help us find 𝑥. 𝐴𝐵 equals eight times the square root of three, 𝐵𝐶 equals log base 𝑥 of 32 plus log base 𝑥 of 1024, 𝐷𝐵 is log base 𝑥 of 1024, and 𝐵𝐴 equals eight times the square root of three. We now have an equation that will ultimately help us find 𝑥. The first stage of this problem involved recognizing some properties of similar triangles. And the second stage is going to be about recognizing properties of logs that allow us to simplify this proportion before we solve. How could we simplify log base 𝑥 of 32 and log base 𝑥 of 1024?
One rule that we have for logs is that log base 𝑏 of 𝑥 to the 𝑎 power is equal to 𝑎 times log base 𝑏 of 𝑥. And that means to try and simplify this log, we want to know, can we rewrite 32 or 1024 as some exponents, some base to some power? As both of these values are even, I want to consider, can I rewrite 32 or 1024 as an exponent with a base of two? In fact, 32 is equal to two to the fifth power. And that means log base 𝑥 of two to the fifth power could be rewritten as five times log base 𝑥 of two. If we’re dividing 1024 by two, you can divide it by two 10 times because two to the 10th power equals 1024. And log base 𝑥 of two to the 10th power can be rewritten as 10 times log base 𝑥 of two.
These values are now as simplified as we can currently take them. So we want to substitute these values in in the equation we’ve already written so that we have eight times the square of three over five times log base 𝑥 of two plus 10 times log base 𝑥 of two is equal to 10 times log base 𝑥 of two over eight times the square root of three. Since we have five log base 𝑥 of two and we’re adding 10 log base 𝑥 of two, we end up with 15 log base 𝑥 of two. From here, our next step will be to cross multiply so that we have eight times the square root of three multiplied by eight times the square root of three is equal to 15 times log base 𝑥 of two times 10 times log base 𝑥 of two. Eight times the square root of three times itself is 192.
On the right side of the equation, we can multiply 15 by 10 to get 150. And then we have log base 𝑥 of two multiplied by itself, log base 𝑥 of two, which will give us log base 𝑥 of two squared. It’s worth pausing here to say that we can’t simplify log base 𝑥 of two squared in the same way that we simplified log base 𝑥 of 32. This is because log base 𝑏 of 𝑥 to the 𝑎 power is being simplified since the base and the exponent are both included in the log. In our case, we are squaring the entire log, not just what’s inside the log function. And that means for now we’ll leave it as is.
Since our ultimate goal is to find 𝑥, we now need to start thinking about how to isolate 𝑥, the first step here being dividing both sides of the equation by 150. 192 divided by 150 written as a fraction in simplest form is 32 over 25. We now have 32 over 25 is equal to log base 𝑥 of two squared. At this point, we can get rid of that square by taking the square root of both sides of the equation so that we have log base 𝑥 of two is equal to the square root of 32 over 25. And now we’re ready to start the third and final stage of solving for 𝑥.
With the equation log base 𝑥 of two is equal to the square root of 32 over 25, we can simplify this square root. 32 equals 16 times two, the square root of 16 is four, so we can simplify the square root of 32 to be four times the square root of two. And the square root of 25 is five. So log base 𝑥 of two is equal to four times the square root of two over five. But we now have something that’s recognizable to the form log base 𝑏 of 𝑥 equals 𝑦, which can be rearranged to say 𝑥 equals 𝑏 to the 𝑦 power. For us, 𝑥 to the four times square root of two over five power is equal to two. This feels like a really strange exponent, and it might not be intuitive how we get from 𝑥 to the four times the square root of two over five power to simply 𝑥.
We need 𝑥 to the first power or just 𝑥. And to get there, we will have to use the rule that 𝑥 to the 𝑎 power to the 𝑏 power is equal to 𝑥 to the 𝑎 times 𝑏 power. We want to raise 𝑥 to the four times the square root of two over five power to some power such that when they’re multiplied together, they equal one. This means we want to raise 𝑥 to the inverse of the power we’re already taking it to, which will be five over four times the square root of two. But if we raise one side of the equation to that power, we need to also raise the other side of the equation to that power. We now have on the left side of the equation 𝑥 to the first power. And on the right side of the equation, we have two to the five over four times the square root of two power, which we can plug into our calculator.
Just make sure when you’re plugging in the power of two, you wanna take two to the five divided by four times the square root of two. Make sure you’re grouping that four times the square root of two together, and you’ll get 𝑥 is equal to 1.8453 continuing, which when rounded to the nearest tenth is 1.8. That means that the segment 𝐷𝐵 will be equal to log base 1.8 of 1024, which is about 11.8 centimeters. But this question was only asking the value of 𝑥, and we’ve determined that 𝑥 rounded to the nearest tenth is 1.8.
