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How do I solve 4^2x3=5  log3 (3x+9)-log3(x-5)=3

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monique06 | Valedictorian

Posted June 6, 2013 at 4:43 PM via web

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How do I solve

4^2x3=5 

log3 (3x+9)-log3(x-5)=3

Tagged with logarithms, math

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted June 6, 2013 at 5:00 PM (Answer #1)

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You need to convert the difference of logarithms of base 3 into the logarithm of quotient, such that:

`log_3 (3x+9) - log_3 (x-5) = 3 => log_3 ((3x + 9)/(x - 5)) = 3`

You need to use the following logarithmic identity, such that:

`log_a b = c => b = a^c`

Reasoning by analogy yields:

`log_3 ((3x + 9)/(x - 5)) = 3 => ((3x + 9)/(x - 5)) = 3^3`

`((3x + 9)/(x - 5)) = 27 `

Reducing duplicate factors yields:

`(x + 3)/(x - 5) = 9 => x + 3 = 9(x - 5) => x + 3 = 9x - 45`

Isolating the members that contain x to the left sides yields:

`x - 9x = -3 - 45 => -8x = -48 => x = 6`

Testing `x = 6` in equation yields:

`log_3 (3*6+9) - log_3 (6-5) = 3`

`log_3 27 - log_3 1 = 3 => log_3 (3^3) - 0 = 3 => 3*log_3 3 = 3`

`3*1 = 3 => 3=3`

Hence, evaluating the solution to logarithmic equation yields `x = 6.`

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