# Video: Using Laws of Logarithms to Find the Area of a Right Triangle

Triangle 𝐴𝐵𝐶 is right-angled at 𝐴. Find its area, given that 𝐴𝐵 = log₂ 49 cm and 𝐴𝐶 = log₇ 512 cm.

03:12

### Video Transcript

Triangle 𝐴𝐵𝐶 is right angled at 𝐴. Find its area, given that 𝐴𝐵 is equal to log to the base two of 49 centimetres and 𝐴𝐶 is equal to log to the base seven of 512 centimetres.

So the first thing I’ve done is that I’ve sketched the triangle. So we’ve got triangle here 𝐴𝐵𝐶. And we were told that it was right angled at 𝐴. So I’ve got 𝐴 at the bottom there by the right angle. And then, we were also told the length of two of the sides. So we’ve got 𝐴𝐵, which is log to the base two of 49 centimetres, and 𝐴𝐶, which is log to the base seven of 512 centimetres. So I’ve also put these onto our triangle.

So what we’re trying to do is we’re trying to find its area. And what we know is that the area of a triangle is equal to a half the base times the height. So we can use this formula to work out the area of our triangle. So therefore, what we’re gonna get is that the area is equal to a half multiplied by log to the base seven of 512 multiplied by log to the base two of 49. Okay, great. So we’ve got this. But how we’re going to multiply our logarithms?

Well, what we’re gonna do is we’re going to change our logarithms so they both got the same base. And to achieve this, what we have is our change of base formula. And that is that log to the base 𝑏 of 𝑥 is equal to log to the base 𝑑 of 𝑥 over log to the base 𝑑 of 𝑏. So then, if we apply this, and what we’re going to do is we’re gonna change both of our logarithms to the base 10. So we don’t need to write the base if we’re doing log to the base 10. So we’re gonna have a half multiplied by log to the base 10 of 512 over log to the base 10 of seven multiplied by log to the base 10 of 49 over log to the base 10 of two.

Well, now, what we want to do is we want to see if our logs can be in fact written as another log raised to an exponent or raised to a power. Well, 512 is the same as two to the power of nine. And 49 is the same as seven squared. So therefore, we can rewrite the area as a half multiplied by log to the base 10 of two to the power of nine over log to the base 10 of seven multiplied by log to the base 10 of seven squared over log to the base 10 of two.

So now, to simplify a bit further, we can use one of our log rules. And that is if we have log to the base 𝑏 of 𝑥 to the power of 𝑦, this is equal to 𝑦 multiplied by log to the base 𝑏 of 𝑥. So therefore, we can get that the area is equal to a half multiplied by nine log to the base 10 of two over log to the base 10 of seven multiplied by two log to the base 10 of seven over log to the base 10 of two. So now, we can see that we can cancel because we can divide through by log to the base 10 of two and also by log to the base 10 of seven. So this is gonna leave us with the area is equal to a half multiplied by nine multiplied by two.

So therefore, we can say that the area is equal to nine centimetres squared.