Question

Solve the following:
${{x}^{x}}=x$

Answer 1

Hint: In order to solve this question, we should know a few logarithmic properties like $\log {{m}^{n}}=n\log m$ and log 1 = 0. Also, we should know how to solve algebraic equations. And while solving the questions, we have to find the maximum number of possible solutions.

Complete step-by-step answer:
In this question, we have to solve the equation ${{x}^{x}}=x$ for x. For that, we know that if a = b, then log a will definitely be equal to log b. So, we will take a log to both sides of the equation. So, we will get,
$\log \left( {{x}^{x}} \right)=\log x$
Now, we know that $\log {{m}^{n}}=n\log m$. So, we can write $\log {{x}^{x}}=x\log x$. Therefore, we get the equation as,
$x\log x=\log x$
Now, we will subtract both sides of the equation by log x. So, we will get,
$x\log x-\log x=\log x-\log x$
Now, we know that the equal terms with opposite sides on the same side of the equation get canceled out. So, we can write the equation as,
$x\log x-\log x=0$
Now, we know that we can take log x common from the left-hand side of the equation. So, we can write x log x – log x as [log x] (x – 1). Therefore, we get the equation as,
$\left( x-1 \right)\log x=0$
Now, we know that for the equation to be 0 either (x – 1) should be 0 or log x should be 0. So, we can write,
x – 1 = 0 and log x = 0
And we know that log x = 0 only when x = 1, so we get,
x – 1 = 0 and x = 1
x = 1 and x = 1
Hence from both the cases, we get the value of x as 1.
Therefore, x = 1 is the only solution of ${{x}^{x}}=x$

Note: In this question, one might think of canceling log x on the LHS with the log x on the RHS which is obviously correct and if we consider this question then it will give the correct answer. But sometimes, in other questions, canceling the common term in the LHS and RHS gives us 50 % of the answer because of which, we can lose marks.