Solve for x, when log 3 ⁡ ( 2 − 3 x ) = log 9 ⁡ ( 6 x 2 − 19 x + 2 )

Question

Solve for x, when       log    3    ⁡  (  2  −  3  x  )  =      log    9    ⁡  (  6      x    2    −  19  x  +  2  )

coquinarq1 · Accepted Answer

Raise both sides to the 9th power to get:

9^{\log_{3} (2 - 3 x)} = 9^{\log_{9} (6 x^{2} - 19 x + 2)}

(3^{2})^{\log_{3} (2 - 3 x)} = 6 x^{2} - 19 x + 2

(3^{\log_{3} (2 - 3 x)})^{2} = 6 x^{2} - 19 x + 2

(2 - 3 x)^{2} = 6 x^{2} - 19 x + 2

This is a quadratic in x. Solve appropriately.
One last thing (thanks to Calvin Lin): Keep in mind that in order for the

\log x

to be defined (over the real numbers), then x>0. After you solve your quadratic, you want to check for extraneous roots. Make sure that your solutions for x satisfy that

2 - 3 x > 0

and

6 x^{2} - 19 x + 2 > 0

(You shouldn't have any problems here, but it's important to do so in general!)

Eliaszowyr1 · Accepted Answer

HINTLet      log    3    ⁡  (  2  −  3  x  )  =      log    9    ⁡  (  6      x    2    −  19  x  +  2  )  =  a    ⟹        9    a    =  6      x    2    −  19  x  +  2 and       3    a    =  2  −  3  x    ⟹        9    a    =  (      3    2        )    a    =  (      3    a        )    2    =  (  2  −  3  x      )    2   So, we have   6      x    2    −  19  x  +  2  =  (  2  −  3  x      )    2  which is a Quadratic Equation of x on re-arrangement

Solve for x, when log 3 </msub> ⁡ ( 2 −<!-- -

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