what to do with: logarithmic, trigonometric and exponential inequalities with variable outside

Question

After encountering this inequality: $$ e^{x/2}=2x+1 $$ that leads me to: $$ x=2\ln(2x+1) $$

I realized that I don't know how to solve it.

But this lack of knowledge expands also to $\cos(x)=x$ or $\ln(x)<\cos(x)$ or $2x=e^x$, $\sin(x)+3\cos(x)>2(x)$ etc. .

I mean is there a way to solve these inequalities (or equations) where the unknown variable is both inside and outside of one of these functions?

p.s. This comes from a function analysis, while I was doing a derivative I found that function. So what to do when you have to study the sign of functions like those?

Answer 1

Equation such as $$e^{x/2}=2x+1$$ does not show analytical solution in terms of elementary functions and, most of the time, numerical methods (such as Newton as Alex Silva commented) should be used.

However, any equation which, after may be some transforms, can write $$a+b x+c\log(d+ex)=0$$ has solutions in terms of Lambert function (http://en.wikipedia.org/wiki/Lambert_W_function) and in your case, beside the trivial $x=0$, the solution write $$x=-2 W_{-1}\left(-\frac{1}{4 \sqrt[4]{e}}\right)-\frac{1}{2}\approx 4.67333$$

Answer 2

Notice that $x=0$ is a solution. Notice also that its derivative has a single root, namely $4\ln2$, being negative for $x<4\ln2$, and positive for $x>4\ln2$, implying that our function is descending for x $<4\ln2$, and ascending for $x>4\ln2$. Since $f(4\ln2)<0$, it follows that we have exactly one root, namely $0$, on the former interval, and exactly one root on the latter. I offer this as a supplement to complete Claude Leibovici's wonderful answer.