# 2e^(x)=3-e^(x+1) what is the value of x in this question?

## can somebody solve this please

Mar 10, 2018

$x = \ln | \frac{3}{e + 2} |$
$x = - 0.4529 = - 0.453$

#### Explanation:

$\textcolor{red}{{a}^{X} = n \iff X = {\log}_{a} n}$
$2 {e}^{x} = 3 - {e}^{x + 1} \implies {e}^{x + 1} + 2 {e}^{x} = 3 \implies {e}^{x} \left({e}^{1} + 2\right) = 3$
${e}^{x} = \frac{3}{e + 2} \iff x = {\log}_{e} | \frac{3}{e + 2} | = \ln | \frac{3}{e + 2} |$
Note:
$x = {\log}_{e} | \frac{3}{e + 2} | = \frac{{\log}_{10} | \frac{3}{e + 2} |}{{\log}_{10} e} ,$....change base.
$x = \left({\log}_{10} | \frac{3}{e + 2} |\right) \cdot {\log}_{e} 10$
$x = \left[{\log}_{10} 3 - {\log}_{10} \left(e + 2\right)\right] \cdot {\log}_{e} 10$
x=[log_10 3-log_10(2.7183+2)*log_e 10
$x = \left[{\log}_{10} 3 - {\log}_{10} 4.7182\right] \cdot {\log}_{e} 10$
$x = \left[0.4771 - 0.6738\right] \cdot \left(2.3026\right)$
$x = \left(- 0.1967\right) \left(2.3026\right)$
$x = - 0.4529 = - 0.453$
Remember: $\textcolor{red}{\left(1\right) e = 2.7183 \ldots \mathmr{and} \left(2\right) {\log}_{e} 10 = 2.3026 \ldots}$
color(red)((3)log_x y=(log_10 y)/(log_10 x)....(to change base)
color(red)((4)1/(log_m n)=log_n m

Mar 10, 2018

Real solution:

$x = \ln \left(\frac{3}{2 + e}\right)$

Complex solutions:

$x = \ln \left(\frac{3}{2 + e}\right) + 2 k \pi i \text{ }$ for any integer $k$

#### Explanation:

Given:

$2 {e}^{x} = 3 - {e}^{x + 1}$

Add ${e}^{x + 1} = e {e}^{x}$ to both sides to get:

$\left(2 + e\right) {e}^{x} = 3$

Divide both sides by $2 + e$ to get:

${e}^{x} = \frac{3}{2 + e}$

We can find the real solution by taking the natural log of both sides to find:

$x = \ln \left(\frac{3}{2 + e}\right)$

To find the complex solutions note that ${e}^{2 k \pi i} = 1$ for any integer $k$. Hence:

${e}^{\ln \left(\frac{3}{2 + e}\right) + 2 k \pi i} = \frac{3}{2 + e} \cdot {e}^{2 k \pi i} = \frac{3}{2 + e}$

So the general solution is:

$x = \ln \left(\frac{3}{2 + e}\right) + 2 k \pi i \text{ }$ for any integer $k$