Question

How do you solve `e^ { 4x - 5} \cdot e ^ { - x } = 4e`?

Answer 1

`x = 1/3 (5 + ln(4e)) ~~ 2.462`

or `x = 2 + 1/3ln4 ~~ 2.462`

Explanation:

Given: `e^(4x-5)*e^(-x) = 4e`

First you must know your rules of exponents: `x^m x^n = x^(m+n)`

Using this rule: `e^(4x-5-x) = 4e; " "e^(3x-5) = 4e`

Natural log both sides of the equation:`" " ln e^(3x-5) = ln 4e`

Since the natural logarithm function `ln x` is the inverse of `e^x`,
`ln e^x = x " (exponent")`.

This means `ln e^(3x-5) = 3x-5`

So, `" "3x - 5 = ln 4e`

Simplifying: `3x = 5+ ln 4e; " " x = 1/3(5+ ln 4e)~~ 2.462`

A second way to solve starting from `e^(3x-5) = 4e`:

Divide by `e`: `" "e^(3x-5)/e = 4`

Rewrite: `" "e^(3x-5)*e^(-1) = 4`

Combine exponents: `" "e^(3x-6) = 4`

Natural log both sides of the equation:`" " ln e^(3x-6) = ln 4`

`ln e^(3x-6) = 3x-6`

So, `" "3x-6 = ln 4`

Simplify: `" " 3(x-2) = ln 4`

`x - 2 = 1/3 ln4`

`x = 2 + 1/3 ln 4~~ 2.462`