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log3(8x+3) = 1+ log3 (x^2)

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clara2 | Student, Undergraduate | (Level 3) eNoter

Posted April 20, 2011 at 11:21 AM via web

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log3(8x+3) = 1+ log3 (x^2)

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted April 20, 2011 at 11:26 AM (Answer #1)

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We have : log3(8x+3) = 1+ log3 (x^2)

log3(8x+3) = 1+ log3 (x^2)

=> log3(8x+3) - log3 (x^2) = 1

use the property log a - log b = log (a/b)

=> log3 [(8x + 3)/(x^2)] = 1

=> (8x + 3)/(x^2) = 3

=> 8x + 3 = 3x^3

=> 3x^2 - 8x - 3 = 0

=> 3x^2 - 9x + x - 3 = 0

=> 3x(x - 3) + 1(x - 3) = 0

=> (3x + 1)(x - 3)= 0

=> x = -1/3 and x = 3

Both the values are defined for the logs given.

The solution of the equation is x = -1/3 and x = 3

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tonys538 | TA , Undergraduate | (Level 1) Valedictorian

Posted October 26, 2014 at 5:33 AM (Answer #3)

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The equation `log_3(8x+3) = 1+ log_3 (x^2)`   has to be solved for x.

`log_3(8x+3) = 1+ log_3 (x^2)`

=> `log_3(8x+3) - log_3 (x^2) = 1`

Use the property log a - log b = log(a/b)

=> `log_3((8x+3)/(x^2)) = 1`

Use the property `log_b b = 1`

This gives: `(8x+3)/(x^2) = 3`

8x + 3 = 3x^2

3x^2 - 8x - 3 = 0

3x^2 - 9x + x - 3 = 0

3x(x - 3) + 1(x - 3) = 0

(3x + 1)(x - 3) = 0

3x + 1 = 0, x = -1/3

and x - 3 = 0, x = 3

Both the values x = -1/3 and x = 3 satisfy the given equation

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