4
$\begingroup$

$$3^x = 3 - x$$

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.

$\endgroup$
  • $\begingroup$ Assuming $x\in\mathbb{R}$, and since $3^x>0$ for all $x\in\mathbb{R}$, then $3-x>0$ which implies $x<3$. $\endgroup$ – Pixel Jul 23 '15 at 8:45
3
$\begingroup$

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

$$ x_0=3-\frac{W(27\ln 3)}{\ln 3}=\color{red}{0.741551813...} $$

$\endgroup$
0
$\begingroup$

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:

enter image description here


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.

$\endgroup$
-1
$\begingroup$

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 $x > 3$.

enter image description here

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.