Question

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

can somebody solve this please

Answer 1

`x=ln|3/(e+2)|`
`x=-0.4529=-0.453`

Explanation:

`color(red)(a^X=n<=>X=log_an)`
`2e^(x)=3-e^(x+1)=>e^(x+1)+2e^x=3=>e^x(e^1+2)=3`
`e^x=3/(e+2)<=>x=log_e|3/(e+2)|=ln|3/(e+2)|`
Note:
`x=log_e|3/(e+2)|=(log_(10)|3/(e+2)|)/(log_10 e),`....change base.
`x=(log_(10)|3/(e+2)|)*log_e 10`
`x=[log_10 3-log_10 (e+2)]*log_e 10`
`x=[log_10 3-log_10(2.7183+2)*log_e 10`
`x=[log_10 3-log_10 4.7182]*log_e 10`
`x=[0.4771-0.6738]*(2.3026)`
`x=(-0.1967)(2.3026)`
`x=-0.4529=-0.453`
Remember: `color(red)((1)e=2.7183...and(2)log_e 10=2.3026...)`
`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`

Answer 2

Real solution:

`x = ln (3/(2+e))`

Complex solutions:

`x = ln (3/(2+e)) + 2kpi i" "` for any integer `k`

Explanation:

Given:

`2e^x = 3-e^(x+1)`

Add `e^(x+1) = e e^x` to both sides to get:

`(2+e)e^x = 3`

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

`e^x = 3/(2+e)`

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

`x = ln (3/(2+e))`

To find the complex solutions note that `e^(2kpi i) = 1` for any integer `k`. Hence:

`e^(ln (3/(2+e)) + 2kpi i) = 3/(2+e) * e^(2kpi i) = 3/(2+e)`

So the general solution is:

`x = ln (3/(2+e)) + 2kpi i" "` for any integer `k`