Can You Solve This Crazy Exponential Equation?

Today we will look at a very interesting exponential equation. Let's consider the following problem; x to the power x cube = 27 to the power of x square. This is a very interesting exponential problem. Let's see how we can solve this problem by using basic algebraic techniques, exponential, and logarithmic properties, as well as some differential calculus. Thank you for viewing this video. Please like the video, and subscribe to our channel if you have not yet done so. Great! Let's see how we can approach this problem. Given that on both sides of the equation, we have exponentials. The best way to start is to take the natural log of both sides. Using the fact that the logarithm and the exponential are inverse functions. We should use the logarithmic function to simplify both sides of this equation. We take the natural log of both sides, this leads to. x cube times log of x equal to x square times log of 27. It is clear that this first operation has significantly reduced the complexity of the problem at hand. Note that 27 is an exponential number. In fact, 27 is 3 to the power 3. Now the equation is, x cube times log of x equals 3x square times log of 3. Then, we move 3x square times log of 3 to the left-hand side, and the problem becomes, x cube times log of x minus 3x square times log of 3 equal to zero. In order to simplify this expression further, we will try to factorize it at most as we can. It is clear that x square is the common factor of both terms of this expression. So we factor x square out and obtain, x square times the quantity x times log of x minus 3 times log of 3. Now given that we have a product of two quantities equal to zero, it means that each of these quantities should be zero. So we end up with two equations; x square equal to zero, and x times log of x minus 3 times log of 3 equal to zero . The first equation x square 0, leads to x equal to zero. But 0 cannot be a solution because the log of zero is undefined. As a matter of fact, each solution should also be in the domain of definition of the natural logarithmic function. Which is the set of positive real numbers. Thus we end up with the last equation x times log of x = 3 times log of 3. With x strictly greater than 0. The trivial solution of x times log of x = 3 times log of 3 is of course x = 3, because with x equal to 3 the left-hand side is exactly equal to the right-hand side. Now, the last question we have to ask ourselves is, are there any other solutions? Or is x=3 the only solution? In order to study the uniqueness of the solution x equal to 3. We define the function f of x. f of x equals x times log of x minus 3 times log of 3. In order to address the uniqueness of solutions, we need to study the variations of the function f of x, thus we have to compute the derivative of f of x. f prime of x equal the derivative of x times log of x plus x times the derivative of log of x. Therefore f prime of x is log of x plus 1. We set the derivative of f equal to zero to find the critical points. So f has only one extremum, f attains its minimum at x equal to exponential of negative 1 This is because its second derivative f second is equal to 1 over x which is always positive for x positive. Therefore f is decreasing from zero to exponential of negative 1 and increasing from exponential of negative 1 to infinity. First, we note that f of exponential of negative 1 is negative exponential of negative 1 minus 3 times log of 3. Then, we compute the limit of f at zero plus and find that the limit of f is negative 3 times log of 3. finally, we compute the limit of f at positive infinity and find that the limit of f is positive infinity. The limits are computed using the hospital rule. In sum, the function f of x only crosses the x-axis at the point x equal to 3. So 3 is the unique solution. Thanks for watching our video. Please smash the like button and subscribe to our channel if you have not yet done so. We have an extensive collection of mathematical videos covering a wide range of topics