e^{2 - 3x}=5 - 3x\nround to the nearest hundredth.\nif there is more than one solution, separate them with…

Explanation:

Step1: Let $y = 2 - 3x$, then the equation becomes $e^{y}=3 + y$.

The original equation $e^{2 - 3x}=5 - 3x$ can be rewritten after substitution.

Step2: Use a graphical or numerical method.

We can graph $f(y)=e^{y}-y - 3$ and find its roots. Or use a numerical - method like Newton - Raphson method. The Newton - Raphson formula for finding the root of a function $f(y)$ is $y_{n + 1}=y_{n}-\frac{f(y_{n})}{f'(y_{n})}$, where $f(y)=e^{y}-y - 3$ and $f'(y)=e^{y}-1$. Let's start with an initial guess. If we consider the function $y = e^{y}-y - 3$, when $y = 1$, $f(1)=e^{1}-1 - 3=e - 4\approx2.718 - 4=-1.282$; when $y = 2$, $f(2)=e^{2}-2 - 3=e^{2}-5\approx7.389 - 5 = 2.389$. Using the Newton - Raphson method with $y_0 = 1.5$: $f(y)=e^{y}-y - 3$, $f'(y)=e^{y}-1$. $y_1=y_0-\frac{e^{y_0}-y_0 - 3}{e^{y_0}-1}$ $y_1 = 1.5-\frac{e^{1.5}-1.5 - 3}{e^{1.5}-1}\approx1.5-\frac{4.482 - 4.5}{4.482 - 1}\approx1.5+\frac{0.018}{3.482}\approx1.505$. $y_2=y_1-\frac{e^{y_1}-y_1 - 3}{e^{y_1}-1}\approx1.505-\frac{e^{1.505}-1.505 - 3}{e^{1.505}-1}\approx1.505-\frac{4.516-4.505}{4.516 - 1}\approx1.505-\frac{0.011}{3.516}\approx1.502$. $y_3=y_2-\frac{e^{y_2}-y_2 - 3}{e^{y_2}-1}\approx1.502-\frac{e^{1.502}-1.502 - 3}{e^{1.502}-1}\approx1.502-\frac{4.499 - 4.502}{4.499 - 1}\approx1.502+\frac{0.003}{3.499}\approx1.503$. Since $y = 2 - 3x$, then $x=\frac{2 - y}{3}$. Substituting $y\approx1.503$ into $x=\frac{2 - y}{3}$, we get $x=\frac{2 - 1.503}{3}=\frac{0.497}{3}\approx0.17$.

Answer:

$0.17$