Question

How do you solve `\ln ( 2x + 2) = 1.6`?

Answer 1

`x = e^1.6/2 - 1`

Explanation:

Firstly, identify that `color(red)ln` is called the natural logarithm.

This means it is a logarithm of a number N with a base `e`.

`log_e N` which can be expressed as `ln_e N`

Like `log_10 N` can be written as `lg N`, `ln_e N` can be written as `ln N` for short.

Therefore, `ln(2x+2) = 1.6`

Now, taking the exponential `e` for both sides, we get:

`color(red)e^ln(2x+2) = color(red)e^1.6`

Note: `e^color(red)(lna)= color(red)a`

`:.e^color(red)ln(2x+2) = color(red)(2x+2)`

Continuing,

`color(red)e^ln(2x+2) = color(red)e^1.6`

`2x+2 = e^1.6`

`2x = e^1.6 - 2` Dividing both sides by 2,

`x = e^1.6/2 - 1`