$$3^x = 3 - x$$
I have to prove that only one solution exists, and then find that one solution.
My approach has been the following:
$$\log 3^x = \log (3 - x)$$
$$x\log 3 = \log (3 - x)$$
$$\log 3 = \frac{\log (3 - x)}{x}$$
And this is where I get stuck. Any help will be greatly appreciated, thanks in advance.
On
Edited version
After $x \cdot \log 3 = \log (3 – x)$ is $x = \frac {\log (3 – x)}{\log 3} $
which is equivalent to $y = x$ and $y = \frac {\log (3 – x)}{\log 3} $
Plotting the above TWO curves on the same graph, you will find they intersect at only one point (i.e. one solution).
Note also that:- When x > 3, the 'y' of the $y = x$ is much higher than the 'y' from $y = \frac {\log |3 – x|}{log 3}$. That means they will never meet when $x > 3$.
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You may consider the function given by $$ f(x)=3^x-(3-x),\quad x \in \mathbb{R}. $$ We have $$ f'(x)=3^x \cdot\ln3+1>0,\quad x \in \mathbb{R}, $$ Thus the function is strictly increasing on $\mathbb{R}$.
We have $$ \begin{align} f(0)&=1-(3-0)=-2<0\\\\ f(1)&=3-(3-1)=1>0 \end{align} $$ then the unique solution $x_0$ is such that $x_0 \in (0,1)$.
You may observe that $$ 3^x=3-x $$ is equivalent to $$ (3-x)\ln 3 \times e^{(3-x)\ln 3}=3^3 \ln 3 $$ then, using a solution of $Xe^X=3^3 \ln 3$ in terms of the Lambert function, we get
$$ x_0=3-\frac{W(27\ln 3)}{\ln 3}=\color{red}{0.741551813...} $$
Unfortunately, finding the solution explicitly is not possible in terms of elementary functions. You'll need to use the Lambert W function.
You have several methods of doing so, one is to simply sketch the graphs and show that they only intersect once as done below:
A more rigorous approach would be to show that $f(x) = 3^x + x - 3$ is a strictly increasing function and show that it attains both negative and positive values.
So $f'(x) = \ln 3\cdot 3^{x} + 1> 0$ for all real $x$, so the function is strictly increasing. Secondly, we have that $f(0) = \text{negative}$ and $f(5) = \text{positive}$ so it crosses the $x$-axis exactly once and hence has only one root.