Question

How do you solve `2 = e^(5x)`?

Answer 1

`x=0.1386(4dp)`

Explanation:

`e^(5x)=2`

taking natural logs of both sides

`5x=ln2`

`=>x=1/5ln2`

`x=0.1386294361`

Answer 2

`x~~0.1386" to 4 dec. places"`

Explanation:

`"using the "color(blue)"laws of logarithms"`

`•color(white)(x)logx^nhArrnlogx`

`•color(white)(x)log_b b=1`

`"take the ln (natural log ) of both sides"`

`ln2=lne^(5x)`

`5xcancel(lne)^1=ln2`

`x=ln2/5~~0.1386" to 4 dec. places"`

Answer 3

`x~~0.139`

Explanation:

`ln` (Natural Log) cancels with base-`e`, so we can take the natural log of both sides. We have

`ln2=cancel(lne)^(5x)`

`ln2=5x`

`x=ln2/5`

`x~~0.139`

Hope this helps!