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

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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 ( 15/2) = log15/2)**

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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