# given that loga (3x-4a)+loga 3x=2/(log2 a)+loga (1-2a),where 0<a<1/2 ,find x

*print*Print*list*Cite

### 1 Answer

You need to use logarithmic identities, hence, you may convert the sum of logarithms from the left into a logarithm of product such that:

`log_a (3x(3x - 4a)) = 2/(log_2 a) + log_a (1 - 2a)`

You need to remember that `log_a b = 1/(log_b a)` such that:

`log_a (3x(3x - 4a)) = 2/(1/(log_a 2)) + log_a (1 - 2a)`

`log_a (3x(3x - 4a)) = 2log_a 2 + log_a (1 - 2a)`

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

You may convert the sum of logarithms from the rigth into a logarithm of product such that:

`log_a (3x(3x - 4a)) = log_a4(1 - 2a)`

Sinc the bases of logarithms are equal, you should equate the numbers such that:

`(3x(3x - 4a)) = 4(1 - 2a)`

Opening the brackets yields:

`9x^2 - 12ax - 4 + 2a = 0`

You need to find x using quadratic formula such that:

`x_(1,2) = (12a +- sqrt(144a^2 - 72a))/18`

You need to put the following conditions for x to exist such that:

`144a^2 - 72a >= 0`

`144a^2 - 72a = 0 => a(144a - 72) = 0`

`a = 0` that is a contradiction since the problem provides th information that `a>0`

`144a - 72 = 0 => a = 72/144 => a = 12/24 => a = 1/2` that is a contradiction since the problem provides th information that `a<1/2`

**Since the problem provides the information that `0<a<1/2` , that means that `144a^2 - 72a< 0=> sqrt(144a^2 - 72a) !in R => x_(1,2) !in R` , hence, there is no real x under the given conditions.**