Question

How do you solve `4ln(5x)+5=2`?

Answer 1

`x = e^(-3/4)/5`

Explanation:

In order to solve this, we will begin with removing 5 from both sides:
`4ln(5x) + 5 - 5 = 2 - 5`.

Thus, we will get `4ln(5x) = -3`.

Then we will divide both sides by 4 to get `ln(5x) = -3/4`.

Then, to get x alone, we need to recognize the relationship of `e` and `ln`. If we rewrite both sides as `e^(ln(5x)) = e^(-3/4)` we can perform our next step.

Remember that `ln` and `e` are inverses of each other and that `ln(e) = 1`. Similarly, `e^ln(x) = x`. If you look at `e^ln(5x)`, you will recognize that we will get `5x` due to `ln` and `e` undoing each other.

Thus, continuing: `5x = e^(-3/4)`.

Divide both sides by 5 and we will get `x = e^(-3/4)/5`.

If you plug this into the equation for x, you will get `4ln(5(e^(-3/4)/5)) + 5 = 2`.

Answer 2

`x = e^(-3/4)/5`

Explanation:

Isolate the logarithm and then cancel it out by raising everything to a common base.

First, subtract 5 from both sides.

`4ln(5x)+5-5=2-5`

`4ln(5x)+cancel5-cancel5=2-5`

`4ln(5x) = -3`

Next, divide both sides by 4.

`4ln(5x) div 4 = -3 div 4`

`cancel4ln(5x) div cancel4 = -3 div 4`

`ln(5x) = -3/4`

Next, raise `e` to the power of both sides. This will cancel out the natural logarithm.

`e^ln(5x) = e^(-3/4)`

`cancele^(cancel"ln"(5x))=e^(-3/4)`

`5x = e^(-3/4)`

And finally divide by 5.

`5x div 5 = e^(-3/4)div5`

`cancel5x div cancel5 = e^(-3/4)/5`

`x = e^(-3/4)/5`

Final Answer