A Harvard interview question|Exponential math problem|can you pass this exam?

You're highly welcome to this channel. So, we are asked that if 45 to the power of x is equal to 75, what is 2x minus 1 divided by x minus 2—that seems somehow tactical, but let's see how this goes went through my mind when I first saw this question was that I could break down 45 into Threes and fives, the same thing with this side, so let me do that first. So to do that, I had to I divided by 3 to get 5, and then I divided by 3 again to get 5. Then I divided by 5 to get 1, which simply means that 45 can be expressed as 3 squared. multiplied by 5, okay, then I'm going to check 75. If I check 75, when 3 divides 7, we have 2, 2 Keep in mind that if we divide 15 by 1, we get 5; then we have 25, and 25 is the same as 5 squared, right? so that means 75 can be expressed as 3 multiplied by 5 squared, so that is exactly what I'm going to do now so 45 to the power of x is 3 squared times 3 squared times 5, okay, everything raised The power of x equals 75 is 3 multiplied by 5 squared. Okay, now I want to expand those brackets. and to do that, recall that a to the power of m multiplied by a a multiplied by b, all raised to the power of a number, is equal to a to the power of x. multiplied by b to the power of x, okay, so that simply means that I can rewrite the left-hand side as 3 to the power of 2x multiplied by 5 to the power of x is equal to 3 multiplied by 5 Squared now I want to bring my threes together and my fives together, okay, and to do that I'm I'm going to divide the left-hand side by 3 first, so let me divide through by 3: divided by 3. divided by 3, now I recall that a to the power of m divided by a to the power of n is equal to a to the power of m minus n, okay, so let's apply that here. If we do that, we have three, we have Three, three, so we are going to pick one three; then we have 2 times the power of three here is one. so it's going to be 2x minus one, okay, multiplied by five to the power of x is equal to now here this we divide this, and we are left with five squared, all right? Now we also, I want to bring five to one side and then three to one side, so I'm going to divide through by what five to the power of x in order to bring it to this side, so I will have three to the power of 2x minus one multiplied by five to the power of x divided by five to the power of x is equal to 1 five to the power of two divided by five to the power of x, okay, so we divide this and we 3^(2x - 1) = 5^(2 - x) I've already explained this particular law all right now at this point we need to introduce log to both sides of the equation, but I'm going to introduce logarithm base three instead, yes, because I have something I want to rewrite this so I have log base three of three to the power of 2x minus one. is equal to the logarithm base three of five raised to the power of two minus x. Now when you have log a to the power of b is equal to b multiplied by log a. All right, so we can now bring down that simply means that we can bring down the powers, we can bring down the powers So we now have 2x minus one log base three of three equal to two minus x log five. base three, all right? Now I want us to note that log a base a is equal to one; therefore, log three three to the power of zero is equal to one, therefore that means the left-hand side of the equation becomes what We have 2x minus one equals two minus x log three base five. Now let's divide both sides. divide this equation by two minus x, so we have divided by two minus x divided by two minus x Now we are going to have 2x minus one divided by two minus x, okay, is equal to this is going to Cancel this. We have log base three of five. All right, I want us to check our question. Let's not be too fast now the question says we should solve 2x minus one divided by x minus two, 2x minus one divided by x minus two, so when you look at this, we have 2x minus one, but we don't have x minus Two here, so that is the question: what do we do to change this? I need to multiply this by minus one. but in mathematics you can't just multiply one side of the equation by minus one; you either multiply through, or multiply both numerator and denominator by the same thing so that nothing changes So what I would do is minus one multiplied by 2x minus one divided by minus one multiplied by two. minus x, that way I haven't really added anything because I can cancel this and this is the same The thing is equal to log base three of five. All right, now I'm going to leave the numerator the way it is. Yes, because if I touch the numerator, I've changed it again, so I'm going to leave it. and change the denominator first, so I will have minus one multiplied by 2x minus one divided by minus one times two gives us minus two. Excuse me, minus one times minus x will give us plus x, all right, so I've gotten the denominator, which says x minus two, right? x minus two, I've Now we have this equal to log base three of five, but we have a problem here—the minus one. What do we do? The only way to clear this is if I multiply both sides of the equation by minus. So I will multiply both sides, that is the left-hand side and the right-hand side, by minus one and if I do that, minus one will cancel minus one, giving us 2x minus one divided, I'm going to write this as x minus two okay is equal to then if I multiply this side by minus one I will have minus log base three of five, and this becomes our final answer because this is actually what we are asked to solve, so it means that the answer is minus log base three of five, so whatever We got log base three of five; we are going to add the negative to it. Thanks so much for watching and