Solving an exponential equation with x as a base and an exponent

Question

So here's the problem:

$x+3=3^x$

Obviously, graphing both sides and finding the intersection would reveal the answer, but algebraically, how can this be solved?

Answer 1

As already said in comments, there are quite many things you can do if you consider the function and its derivatives $$f(x)=x+3-3^x$$ $$f'(x)=1-3^x \log (3)$$ $$f''(x)=-3^x \log ^2(3)$$ The first derivative cancels for $$x_*=-\frac{\log (\log (3))}{\log (3)}\approx -0.085606$$ For this value $$f(x_*)=3-\frac{1}{\log (3)}-\frac{\log (\log (3))}{\log (3)}\approx 2.00415$$ and the second derivative is always negative; so, the equation shows two roots.

Graphing the function, you certainly noticed that one root is close to $-3$ and the other one close to $1.5$. This is enough to start a simple root-finding method such as Newton; starting from a "reasonable" guess $x_0$, the method will update it according to $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ Let us apply it to the first root starting using $x_0=-3$; the iterates will be $$x_1=-2.96139$$ $$x_2=-2.96136$$ which is the solution for six significant figures.

Let us apply it to the second root starting using $x_0=1.5$; the iterates will be $$x_1=1.35215$$ $$x_2=1.33529$$ $$x_3=1.33509$$ which is the solution for six significant figures.

Also as mentioned in comments, you will (sooner or later) that any equation which can write or rewrite $A+Bx+C\log(D+Ex)=0$ has solutions in terms of Lambert function. For the time being, I shall not go further; at the time you will be ready for it, let me know and I shall continue.

Answer 2

$$x+3=3^x$$ The equation has two real solutions which can be computed by numerical calculus. If you have a mathematical sofware with the Lambert W function, the analytical solution is :

$$x+3=\frac{3^{x+3}}{3^3}=\frac{ e^{ (x+3)\ln(3) } }{27}$$ $$-(x+3)\ln(3)e^{-(x+3)\ln(3)}= -\frac{\ln(3)}{27}$$ With $Y=-(x+3)\ln(3)$ and $X=-\frac{\ln(3)}{27}$ $$Ye^Y=X\quad \to \quad Y=W(X)$$ $W(X)$ is the Lambert's W function. $$-(x+3)\ln(3)=W\left(\frac{-\ln(3)}{27}\right)$$ $$x=-3-\frac{1}{\ln(3)}W\left(-\frac{\ln(3)}{27}\right)$$ The Lambert $W(X)$ real function is multivaluated on the negative range $-e^{-1}<X<0$

So, they are two solutions usually noted $W_0(X)$ and $W_{-1}(X)$ leading respectively to : $$x\simeq \begin{cases} -2.96135674004667\\1.33508549596605 \end{cases}$$