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What is the answer for question 16) ? http://postimg.org/image/6g52noj6f/

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lkballer24 | Student, Grade 11 | Valedictorian

Posted May 28, 2013 at 11:34 PM via web

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What is the answer for question 16) ?

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Mary Joy Ripalda | High School Teacher | (Level 3) Educator

Posted May 29, 2013 at 2:35 AM (Answer #1)

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`log_3x+log_3(x+2)=1`

To explain this through the telephone, we can describe the steps as follows:

  • Notice that the two logarithms in the equation have the same base. So, express it as one logarithm using the product property.
  • So, at the left side of the equation write log_3 and multiply the two arguments x and (x+2). And the right side remains the same.

`log_3(x*(x+2)) = 1`

  • Since `x*(x+2` ) is equal to `x^2+2x` , the argument of the logarithm becomes `x^2+2x` .

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

  • Since the logarithm of  `x^2+2x` is equal to 1, then we apply the property that a logarithm is equal to 1 if its base and argument are the same `(log_b b=1)` .
  • This means that the argument of `log_3` is equal to 3. So, set `x^2+2x` equal to 3. And this becomes our new equation.

`x^2+2x = 3`

  • Now that we have a quadratic equation, to solve for x, set one side equal to zero. To do this, subtract both sides by 3.

`x^2+2x - 3 = 3-3`

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

Then, factor left side.

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

Next, set each factor equal to zero.

`x+ 3= 0`     and     `x-1=0`

So the equation breaks into two.

For the first equation, subtract both sides by 3.

`x+3=0`

`x+3-3=0-3`

`x=-3`

Note that in logarithm, we can not have a negative number as its argument. So, we do no consider -3 as a solution to our equation.

Next, solve for x in the second equation. To do so, add both sides by 1.

`x-1=0`

`x-1+1=0+1`

`x=1`

Hence, the solution to the given equation` log_3x + log_3(x+1)` is `x=1` .

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