SOLUTION: Solve algebraically: log2(2-2x) + log2(1-x)=5 Solve algebraically: 2log(3-x)= log4 + log(6-x) Solve 2^x = 3(5^x+1)

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Question 1206970: Solve algebraically: log2(2-2x) + log2(1-x)=5


Solve algebraically: 2log(3-x)= log4 + log(6-x)


Solve 2^x = 3(5^x+1)

Answer by ikleyn(52786)   (Show Source): You can put this solution on YOUR website!
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Solve algebraically: log2(2-2x) + log2(1-x)=5
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Due to properties of logarithms, this equation is equivalent to 

    (2-2x)*(1-x) =   in the domain  1-x > 0.


Simplify the last equation

    2 - 2x - 2x + 2x^2 = 32,   

    2x^2 - 4x - 30 = 0,

    x^2 - 2x - 15 = 0

    (x-5)*(x+3) = 0.

The solutions to the last equation are x= 5 and x= -3;  but only x= -3 is in the domain.


ANSWER.  The only solution is x = -3.

Solved.


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Solve algebraically: 2log(3-x)= log4 + log(6-x)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Due to properties of logarithms, this equation is equivalent to

    (3-x)^2 = 4*(6-x)  in the domain  x < 3.


Simplfy the equation

    9 - 6x + x^2 = 24 - 4x

    x^2 -2x - 15 = 0

    (x-5)*(x+3) = 0.


The solutions to the last equation are x= 5 and x= -3;  but only x= -3 is in the domain.


ANSWER.  The only solution is x = -3.

Solved.



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