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log 3x^2 - log x = log (x+5)    find x

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seedrasoso | Student, Grade 10 | eNoter

Posted November 14, 2010 at 6:22 AM via web

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log 3x^2 - log x = log (x+5)    find x

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hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted November 14, 2010 at 6:29 AM (Answer #1)

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log 3x^2 - log x  = log (x+5)

First we will use the algorethim proprties  to simplify the equation.

We know that: log a - log b = log a/b

==> log (3x^2)/x  = log (x+5)

Now simplify :

==> log 3x = log (x+ 5)

Now, we know that:

if log a = log b   Then,    a = b

==>3x = x+ 5

==>3x - x = 5

==> 2x = 5

==> x= 5/2

To check :

log (3x^2) - log x = log (x+ 5)

log (3*25/4) - log 5/2 = log (5/2 + 5)

log (75/4) - log(5/2) = log (15/2)

log (75/4) / (5/2)  = log (15/2)

log (75/10) = log (15/2)

log ( 15/2) = log15/2) 

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neela | High School Teacher | Valedictorian

Posted November 14, 2010 at 12:26 PM (Answer #2)

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To solve log3x^2-logx = log(x+5).

log3x^2= log3+logx^2 = log3+2logx, as loga^m = mloga and log(ab) = loga+logb.

Therefore  log3x^2 =  log3 +2logx-log x = log(x+5).

log3 +logx= log(x+5).

log3 = log(x+5)-logx.

log3 = log{(x+5)/x.

We take antilogarithms of both sides:

3 = (x+5)/x.

3x = x+5.

3x-x = 5

2x= 5.

x = 5/2.

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