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logx(8x-3)-logx 4=2

Guest Sep 2, 2017
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 #1
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Solve for x:
(log(8 x - 3))/(log(x)) - (log(4))/(log(x)) = 2

Bring (log(8 x - 3))/(log(x)) - (log(4))/(log(x)) together using the common denominator log(x):
(log(8 x - 3) - log(4))/(log(x)) = 2

Multiply both sides by log(x):
log(8 x - 3) - log(4) = 2 log(x)

Subtract 2 log(x) from both sides:
-log(4) - 2 log(x) + log(8 x - 3) = 0

-log(4) - 2 log(x) + log(8 x - 3) = log(1/4) + log(1/x^2) + log(8 x - 3) = log((8 x - 3)/(4 x^2)):
log((8 x - 3)/(4 x^2)) = 0

Cancel logarithms by taking exp of both sides:
(8 x - 3)/(4 x^2) = 1

Multiply both sides by 4 x^2:
8 x - 3 = 4 x^2

Subtract 4 x^2 from both sides:
-4 x^2 + 8 x - 3 = 0

The left hand side factors into a product with three terms:
-(2 x - 3) (2 x - 1) = 0

Multiply both sides by -1:
(2 x - 3) (2 x - 1) = 0

Split into two equations:
2 x - 3 = 0 or 2 x - 1 = 0

Add 3 to both sides:
2 x = 3 or 2 x - 1 = 0

Divide both sides by 2:
x = 3/2 or 2 x - 1 = 0

Add 1 to both sides:
x = 3/2 or 2 x = 1

Divide both sides by 2:

 x = 3/2     or      x = 1/2

Guest Sep 2, 2017
 #2
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+1

 

logx(8x-3) - logx 4=2         Note that we can write

 

logx ([ 8x - 3] / 4)   = 2   In exponential form, we have that

 

x^2  =  ([ 8x - 3] / 4)       Multiply both sides by  4

 

4x^2  =  8x - 3              rearrange as

 

4x^2 - 8x + 3  = 0        factor as

 

(2x -  3) (2x - 1)  = 0

 

Setting each factor to 0 and solving for x gives us the possible solutions of  x = 3/2  or x = 1/2

 

Checking  x = 3/2   using the change of base method

 

log ( 9) / log (3/2)  - log (4) / log(3/2)   = 2

 

 

 

Checking  x = 1/2   using the change of base method

 

log ( 1) / log (1/2)  - log (4) / log(1/2)   =  2

 

So both solutions are valid

 

 

 

 

cool cool cool

CPhill  Sep 2, 2017

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