Help me, solve 3^(1+log(2, 1/x))-9^(log(4, 4x^2))-11+3^(log(2, 8x^3))=0

1 Answer | Add Yours

sciencesolve's profile pic

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

You need to change the base of logarithm `log_4 (x^2) ` such that:

`log_4 (x^2) = (log_2 (x^2))/(log_2 4)`

Using logarithmic identities yields:

`log_4 (x^2) = (2log_2 x)/(2log_2 2) = log_2 x`

Hence, you may substitute `1 + log_2 x`  for `log_4(4x^2) = log_4 4 + log_4 x^2`  such that:

`3^(1 + log_2 1/x) - 9^(1 + log_2 x) - 11 + 3^(log_2(8x^3)) = 0`

You need to convert the logarithm of product `log_2(8x^3)`  into the sum of logarithms `log_2 8 + log_2 x^3 = 3 + 3log_2 x`  such that:

`3^(1 + log_2 1/x) - 9^(1 + log_2 x) - 11 + 3^(3 + 3log_2 x) = 0`

`3^(1- log_2 x) - 3^(2(1 + log_2 x)) - 11 + 3^(3(1 + log_2 x)) = 0`

You should come up with the substitution `log_2 x = t`  such that:

`3^(1 - t) + 3^(2(1+t)) - 11 + 3^(3(1 + t)) = 0`

`3/3^t + 3^2*3^(2t) - 11 + 3^3*3^(3t) = 0`

You need to multiply by `3^t`  such that:

`3 + 9*3^(3t) - 11*3^t + 27*3^(4t) = 0`

You need to substitute  u for `3^t`  such that:

`3 + 9u^3 - 11u + 27u^4 = 0`

`27u^4 + 9u^3 - 11u + 3 = 0`

You need to try to find the roots among the following values of the set `(D_3)/(D_27)`

`D_3 = {+-1 ; +-3}`

`D_27 = {+1 ; +-3 ; +-9 ; +-27}`

`(D_3)/(D_27) = {+-1/3 ; +-1/9 ; +-1/27}`

`u= 1/3 => 27/81 + 9/27- 11/3 + 3 = 0`

`1/3 + 1/3 - 11/3 + 3 = -9/3 + 3 = 0`

Notice that `u = 1/3`  is a root for `27u^4 + 9u^3 - 11u + 3 = 0`

`u = -1/3 =>27/6561 + 9/729- 11/9 + 3 != 0`

`(27 + 81 - 8019 + 19683)/6561 != 0`

Hence, you may find t solving the equation `3^t = u => 3^t = 1/3 ` =>`t = -1` .

You may find x solving the equation  `log_2 x = t`  such that:

`log_2 x = -1 => x = 1/2`

Notice that the equation has two real roots and two complex roots.

Hence, evaluating one real solution of the equation yields `x = 1/2.`

We’ve answered 318,916 questions. We can answer yours, too.

Ask a question