I was able to solve $ \ln x = - x $ using Lambert's function.  I was wondering how does one solve $ \ln x = x$   Does the solution even exist for this equation?
122107
I am trying to find x values for points along the normal distribution curve, and I ended up with a problem that goes back to the method of solving $x = \ln x$. Right now, I have $\ln(a \mu) - \ln(10) = \frac {-1} {2a^2} $. This is not something I know how to solve. I tried Googling this type of problem and nothing came up. How does one solve for a variable both inside and outside of a natural logorithm, specifically in the example problem above? Could anyone explain how this could be done?
27302
I am so confused,  how to get a solution of following differential equation:  $$(y')^2 -xy' + y = 0$$  My Thought:  Try solving $y'$ to get $y' = \dfrac{ x \pm \sqrt{x^2 - 4y}}{2}$ and now its unsolvable. Kindly help, appreciated.
30870
I plotted the graphs of $x$ and $\ln(x+1)$ and found they meet when $x= 0$, is there any way to show this analitically? I tried $e^x = x+1$ but its actually the same
7879
I'm trying to solve this equation:  $$2-x=-\sqrt{x}$$ Multiply by $(-1)$: $$\sqrt{x}=x-2$$ power of $2$: $$x=\left(x-2\right)^2$$ then: $$x^2-5x+4=0$$ and that means: $$x=1, x=4$$    But $x=1$ is not a correct solution to the original equation.  Why have I got it? I've never got a wrong solution to an equation before. What is so special here?
21786
I need to find $X$ from   $$\begin{pmatrix} 1 &amp; 2\\  -3 &amp;-6  \end{pmatrix} X \begin{pmatrix} 1 &amp;2 \\  -1 &amp;-2  \end{pmatrix}=\begin{pmatrix} 2 &amp;4 \\  -6 &amp; -12 \end{pmatrix}$$  I wrote $X$ as   $$X=\begin{pmatrix} a &amp; b\\  c &amp;d  \end{pmatrix}$$  and I got $a+2c-b-2d=2$ but I do not know what to do next. Please help.
83049
I am looking for a close form solution for below equation.  $$\log_{2}{x}+\log_{3}{x}+\log_{4}{x}=1.$$   I solve it by graphing, but I don't know is there a way to find $x$ analytically  ?
107208
$\log x =Cx^4$ has only one root. Find C.  I don't know how to solve this problem. Do you take derivative on both sides?  I am thinking C equals 0. Am I correct on that?
85684
Find the general solution of $xy''-(1+x)y'+y=x^2$ knowing that the homogeneous equation has the following solution: $e^{ax}$, where $a$ is a parameter you have to find.  I have found that $a=1$ or $a=1/x$.
3511
$$e^x − 6e^{-x} − 1 = 0$$  No idea how to solve this. If someone could show me the first one or two steps to push me in the right direction that would be great.
84329
Solve for   $$ \lim_{x \rightarrow \infty} \frac{x^{\ln(x)}}{(\ln(x))^x} $$   I used l'Hôpital's rule but in vain. Could you help me someway. Thanks for attention.
8644
$$y' = {y \over {\sin x}}+\tan{x \over 2}$$  I was trying to do this by substitution $u=y/x $ and it did not work and also with $$y' - {y \over {\sin x}} = 0$$ $${dy \over dx} = {y \over {\sin x}} $$ $${\ln(y) = \ln\left|\tan\frac{x}{2}\right|+c } $$ $${y = c\cdot\tan\frac{x}{2} } $$ but then when im trying to calculate $y'$ I have a problem and I have too many equations. Is there some easier way or am I making some mistakes.
51127
Solve the following equation:  \begin{array}{l}{\sqrt{\sin x-\sqrt{\cos x+\sin x}}=\cos x} \\ \text{my try as follows:}\\{\sin x-\sqrt{\cos x+\sin x}=\cos ^{2} x} \\ {\sin x-\cos ^{2} x=\sqrt{\cos x+\sin x}} \\ {\sin ^{2} x+\cos ^{4} x-2 \sin x \cos ^{2} x=\cos x+\sin x} \\ {\sin ^{2} x+\cos ^{4} x-2 \sin x \cos ^{2} x-\cos x-\sin x=0} \\ {\sin ^{2} x+\cos ^{4} x-2 \sin x\left(1-\sin ^{2} x\right)-\cos x-\sin x=0} \\{\sin ^2}x + {\left( {1 - {{\sin }^2}x} \right)^2} - 2\sin x\left( {1 - {{\sin }^2}x} \right) - \cos x - \sin x = 0\\ {\sin ^{2} x+\sin ^{4} x-2 \sin ^{2} x+1-2 \sin x+2 \sin ^{3} x-\cos x-\sin x=0} \\ {\sin ^{4} x+2 \sin ^{3} x-\sin ^{2} x-3\sin x-\cos x+1=0}\end{array}  Now i think it gets more complicated , any help would be appreciated
38573
Find all solutions to $$xy=x+y$$ Initially the given condition was $x,y\in \Bbb{Z}$.    $$$$In this case, I just guessed that the solutions were $(0,0)$ and $(2,2)$. As far as I can see, these are the only 2 integral solutions possible. However, i'm quite surprised as usually a Diophantine Equation has infinite solutions. Could somebody please show me how to actually $solve$ this equation instead of just guessing the values? Is there any way to $show$ that there are just 2 sets of solutions?  $$$$Secondly, what if $x,y\in \Bbb{R}$? In that case how could the equation be solved? $$$$For both these conditions on the values of $x,y$, is there any way to use Coordinate Geometry to achieve an answer? $$$$ Many thanks in anticipation!  Edit: I forgot to mention that I know the solution using factoring. I was hoping to find a solution using Coordinate Geometry. I was told by a friend that the non integral solutions lie on 2 unique lines, and am hence particularly interested in a geometrical solution.
98533
Recently saw a question that asked me to show that $\ln\left(\sqrt{2} - 1\right) = - \ln\left(\sqrt{2} + 1\right)$. How can I demonstrate that the LHS equals the RHS?
105982
Find all $x \in (-1, +\infty)$ such that $(x^2 + 4x + 3)^x + (2x + 4)^x = (x^2 + 4x + 5)^x$.  What I have done so far was a substitution $y = x + 2$ which results in a nicer form of the equation:   $(y - 1)^{y - 2}(y + 1)^{y - 2} + 2^{y - 2}y^{y - 2} = (y^2 + 1)^{y - 2}$ and $y &gt; 1$, which can then be rewritten as:  $$ \left( \frac{y^2 + 1}{2y} \right)^{y - 2} - \left( \frac{y^2 - 1}{2y} \right)^{y - 2} = 1 $$ Also, my intuition is that $y = 4$ is the unique solution and initially I wanted to prove that this function of $y$ is injective, but it didn't work either.  Do you have any suggestion on how I can continue?
112459
It recently occurred to me that I did not know how to solve equations of the form $k^x=x^c$ for any two constants $k$ and $c$. After much pain in algebraically manipulating the equation (using logarithms) I confirmed the x is trapped in an exponent and cannot be extracted without some external knowledge.  Finding the roots of the derivative creates a similar problem.   I then did some research and discovered Lambert's W function, which I haven't the mathematical maturity to understand deeply. ().   I'm looking for some rigorous intuition towards why without numerical methods or Lambert's W function the equation is not solvable for x, or more specifically, a proof as to why this class of function is unsolvable using only logs and rationals? The best answer I have for myself is that because x is both a power and an exponent, either extracting x with a log or removing the exponent with a rational will keep the other side of the equation unsimplified. What is the mathematical language with which to express this idea?  As an aside, I'm also wondering what area of mathematics considers issues such as these (this one being on the simpler side).
65016
A certain physics problem I have been working on has turned into a math problem. Particularly, I want to find the solutions of some equation of the form   $$f(x)=ae^{bx}+cx+d = 0$$  where $a, b, c,$ and $d$ are constant, real numbers that come from the physics problem and $x$ will be a real number. I do not know how to find the solutions of an equation of this form. In the special case where $c=0$, the solution is simply  $$x = \frac{\ln(\frac{-d}{a})}{b}$$  but for other values of $c$, I am stumped. If there is no exact form for the roots, is there a relatively simple way to estimate the roots?