Question

How do you solve `ln(2x^2-2)-ln9=ln80`?

Answer 1

The Soln.`: x=+-19`.

Explanation:

`ln(2x^2-2)-ln9=ln80`

`:. ln(2x^2-2)=ln9+ln80=ln(9*80)`

Since, `ln` fun. is `1-1`, we have,

`2x^2-2=9*80`

`rArr 2(x^2-1)=9*80`

`rArr x^2-1=9*40=360`

`rArr x^2=361`

`rArr x=+-sqrt361=+-19`

`x=+19, and, x=-19` satisfy the given eqn.

Hence, the Soln. `x=+-19`.

Answer 2

`x = +-19`

Explanation:

If the logs are being subtracted, the numbers are being divided.

We can condense two ln terms into one.

`ln(2x^2-2) - ln9 =ln80`

`ln((2x^2-2)/9) = ln 80`

`:. (2x^2-2)/9 = 80" if "ln(a/b) = ln c rArr a/b = c`

`2x^2-2 = 720`

`2x^2= 722`

`x^2 = 361`

`x= +-sqrt361`

`x = +-19`