How solve the equation log base 3 (2x^2-3x+1)=0? 

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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By logarithm definition, yields:

`log_a x = b => x = a^b`

Reasoning by analogy, yields:

`log_3 (2x^2-3x+1) = 0 => 2x^2-3x+1 = 3^0`

Since `3^0 = 1` , yields:

`2x^2-3x+1 = 1 => 2x^2 - 3x + 1 - 1 = 0`

Reducing duplicate members, yields:

`2x^2 - 3x = 0`

Factoring out x yields:

`x(2x - 3) = 0`

Using the zero product rule yields:

`{(x = 0),(2x - 3 = 0):} => {(x = 0),(x = 3/2):}`

You need to test the values above in equation, such that:

`x = 0 => log_3 (2*0^2-3*0+1) = log_3 1 = 0 => 1 = 3^0 => 1 = 1`

`x = 3/2 => log_3 (2*9/4-3*3/2+1) = log_3(9/2-9/2+1) = log_3 1 = 0 => 1 = 1`

Hence, testing both values in equation, the equation holds, thus, the solutions to the given equation are `x = 0` and `x = 3/2.`

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user4825270 | Student | (Level 3) eNoter

Posted on

log base 3(2x^2-3x+1)=0
3^0=2x^2-3x+1
1=2x^2-3x+1
2x^2-3x+1-1=0
2x^2-3x=0
x(2x-3)=0
x=0 
  OR
2x-3=0
2x=3
x=3/2

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