Question

Question #db5cd

Answer 1

`1 - lnroot(3)(2) = x`

Explanation:

Use `a^-n = 1/a^n`:

`1/e = 2e^(3x - 4)`

Cross multiply:

`1 = 2e^(3x - 4)e^1`

Use `a^m * a^n = a^(m + n)`

`1 = 2e^(3x - 4 + 1)`

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

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

Take the natural logarithm of both sides.

`ln(1/2) = ln(e^(3x- 3))`

Use `lna^n = nlna`.

`ln(1/2) = (3x- 3)lne`

`ln` and `e` are inverses--their product is `1`

`ln(1/2) = 3x- 3`

`1/3(ln(1/2) + 3) = x`

Use `ln(a/b) = lna - lnb`.

`1/3(ln1 - ln2 + 3) = x`

We know that `ln1= 0`.

`1/3(3 - ln2) = x`

`1 - 1/3ln2 = x`

Use `alnn = lnn^a`.

`1 - ln2^(1/3) = x`

`1 - lnroot(3)(2) = x`

Hopefully this helps!