Omforme

In t And the rules that are valid for the links, we will get t Of course we want to show that t At least that's what happens. So we can try it. It's a pretty good exercise on algebra. And it is often possible to say that you have come up with something And then in the face it is on your form here. And you do not know if you have got the right answer or not. But if they are equivalent, they are the right answer. So it can be great to get an exercise with such an algebra. The first t So if you try to combine them, you can get the opportunity to simplify it. If you divide it into 5 in the 3rd on both sides, then on the left side you just want to stand again with t So x over 5 divided by lnx plus 2. If you divide it into 5 in the 3rd on t And now we can use the rule that says that if you have a break with two potentials with the same exponent, then you can write it as you just pull the exponent out of the 1 bar nth and then you have a break made up of the root n So t And then we have the same root n But now it would be pretty easy to find solutions to these problems. Because the two potentials are the same, and we also see that the root n Then the exponent is also the same, so a solution would be to just set the exponent to be the same as the other one. It would then give x equal to 1, x equal to e. So that would have been a solution on the links, it looks Well, I'm going to give a solution that would be x equal to 5, because if you have 5 here and 5 here, then you get 1 up in the 1th row, w But again, that's all good, we're going to train on. And what strikes me here is that we have potentials, it can be weird to look at the potentials. What we can do here is to take the natural logarithm on both sides. If we just do a little more space here, we have the logarithm here too. Then we can use the rule that says that if you have the logarithm t1 on the left, then it's the same as the exponent x logarithm t prime in the potential. W And the addition here, the exponent, is 3 times the logarithm t prime, w And the goal here is to get a value where you have 0 one side, so we can now pull from then we get ln x plus 2 times ln tx over 5 minus 3 ln x over 5. And then we're just standing here with 0 on the right side, and we've moved then the next step can be. . . Yes, we can perhaps notice that both sides have the factor ln tx over 5, w And then we're standing here with. . . OK, then we have the far side of d and d. We have factorized out d, then we're just standing here with ln x plus 2 minus 3, equal to 0. So with other words, logarithm tx over 5 times logarithm tx, and then it becomes minus 1 here, equal to 0. And then, we're almost in the goal, we've got ln x minus 1, w The other factor is already here, just on a different form. Now it'so we have logarithm tx, i. e. the count minus logarithm for an event. times logarithm tx minus 1 is equal to 0, and that was the goal, that was the purpose of your form, where we wanted to write you. I'm complaining if my voice was a bit monotonous, it's usually a bit common, but maybe a bit extra today, because I'm trying to fight a misuse and a misuse. So if that was a problem, I'm complaining, but I hope that you got a bit out of the video.