Olympiade, Maths

2 exponents m equals m exponents 32. Here m is an exponent and here m is the base. Normally, we have to use the logarithmic function Néperien to bring down the m. But don't worry, we're not going to use the logarithmic function Néperien this time. What do we do? We just have to eliminate the m exponents by multiplying the exponents by the inverse of m. That is, we first put 2 exponents m which is equal to m exponents 32. And there I multiply the exponent here by 1 over m and the exponent here also by 1 over m. Start and others. So I haven't changed anything. So I'll have 2 here and m will multiply 1 over m. So m times 1 over m which is equal. And here we will have m exponents 32 times 1 over m. So here I simplify m and I end up with 2 exponents. 1 gives us 2 which is equal to m. And here 32 times 1 over m gives us 32 over m. That is, I have grouped the m on one side. Now I have to group the 32 with the 2 on the other side. That is, I multiply the exponent of each by the inverse of 32. That is, here 2 exponents 1 over 32 is equal to m exponents 32 over m. I also multiply by 1 over 32. So I get here 2 exponents 1 over 32 is equal to m exponents 32 over m times 1 over 32 in this way. So here I simplify the 32 by 32. I get 2 exponents 1 over 32 is equal to m exponents 1 over m. Now here the base is 2 and the exponent is 1 over 32. Here the base is 1 and the exponent is 1 over m. So that means that the base and the inverse of the exponent must be the same number. That is, I have to transform the quantity here so that the number at the base with the number at the denominator must be the same. So what do I do? I multiply the numerator and the denominator of 1 over 32 by 2. So I'm going to have here 2 exponents 2 times 1 over 32 and I also multiply the 32 by 2. That is, if I multiply the numerator and the denominator of a fraction by the same number, it does not change. So I multiply the numerator by 2 and the denominator by 2 which is equal to m exponents 1 over m. So the 2 that is here, I can say that when you have a to the power of m times n, you can write it as a to the power of m times n. It's the same thing. So that means that here, what are we going to have? We're going to have 2 exponents 2 times 1 over 32 times 2, that's 64, right? Is equal to m exponents 1 over m in this way. So I do it here. Now what do I do? Now 2 squared. 2 squared is equal to 4. So I get 4 exponents 1 over 64 which is equal to m exponents 1 over m. m exponents 1 over m. So now I multiply the numerator and the denominator again. Because the base and the denominator here are not the same yet. I always multiply by 2 until I get a number in the base and in the denominator of the exponents that are the same. I multiply again by 2 until I get 4 exponents 2 times 1, right? So the 2 here is the denominator's 2. The denominator over 2 times 64. Like that, it's equal to m exponents 1 over m. So here, the 2 is going to come in like this, as we did here. So 4 squared is equal to m exponents 1 over m. And 4 squared is equal to 16. So I find 16 exponents 1 over 128 which is equal to m exponents 1 over m. I continue. I multiply the numerator and the denominator by 2. So I do it here. So if I multiply the numerator and the denominator by 2, that is, I will have here 16 exponents 2 times 1 over 128. But the 128 also I multiply by 2, right? The denominator also by 2. The numerator and the denominator by 2. So which is equal to m exponents 1 over m. And here I will have 16 exponents 2, like this, and the denominator by 1 over 2 times 128. 2 times 128 gives us 256, which is equal to m exponents 1 over m, like this. Now what do we do? 16 squared gives me 256. So 256 exponents 1 over 256 which is equal to m exponents 1 over m. You see? So I found the base and the denominator of the exponents which are the same. The base and the denominator of the exponents which are the same. That is, here I put the base. Here m is equal to 256. And 1 over m. So I equalize the bases between them and the exponents between them. 1 over m is equal to 1 over 256. So m is equal to 256. 1 over m is equal to 256. That is, when we subtract m here too, we find 256. So the set of solutions is 256. That is, if you replace m here by 256, m here by 256, you actually get the same value.