help? ln (8-x^3) - ln (2-x) = ln (2x=5)?

Question

any help would be welcome! thank you!

Answer 1

ln(8 - x^3) - ln(2 - x) = ln(2x + 5)

First, combine the logs on the left hand side.

ln( (8 - x^3)/(2 - x) ) = ln(2x + 5)

Take the antilog of both sides, to get rid of the logs.

(8 - x^3)/(2 - x) = 2x + 5

8 - x^3 = (2x + 5)(2 - x)
8 - x^3 = 4x - 2x^2 + 10 - 5x
8 - x^3 = -2x^2 + 10 - x
0 = x^3 - 2x^2 - x + 10 - 8
0 = x^3 - 2x^2 - x + 2

This can easily be solved by grouping.

0 = x^2(x - 2) - (x - 2)
0 = (x - 2)(x^2 - 1)
0 = (x - 2)(x - 1)(x + 1)

Therefore,
x = {2, 1, -1}

However, we have to check for extraneous solutions by plugging each potential solution back into the original equation. Note: we cannot take the log of 0 or a negative number.

x = 2 automatically fails because we have ln(2 - x), which forces us to have ln(0).

Our only solutions are x = {1, -1}

Answer 2

ln (8-x^3) - ln (2-x) = ln (2x-5)
applying the logarithm rule, you'll get:
ln [(8-x^3)/(2-x)] = ln (2x-5)
(8-x^3)/(2-x) = (2x-5)
cross multiply the equation:
(8-x^3) = (2x-5)(2-x)
8 - x^3 = 4x - 2x^2 - 10 + 5x
-x^3 + 2x^2 -5x - 4x +10 +8 = 0
-x^3 + 2x^2 - 9x + 18 = 0
Multiply throughout by -1:
x^3 -2x^2 + 9x - 18 = 0

Then you solve the equation by factorising the equation.

Answer 3

ln (8 - x^3) - ln (2 - x) = ln (2x + 5)?
(8 - x^3) / (2 - x) = (2x + 5)
x^2 + 2x + 4 = 2x + 5
x^2 - 1 = 0
(x + 1)(x - 1) = 0
x = (-1,1)

Answer 4

I think lnx-lny=lnx/y

so ln(8-x^3)-ln(2-x) would equal:
ln[(8-x^3)/(2-x)]