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What is the answer for question 16) ? http://postimg.org/image/6g52noj6f/
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- 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.
- Since `x*(x+2` ) is equal to `x^2+2x` , the argument of the logarithm becomes `x^2+2x` .
- 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.
- 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.
High School Teacher
To explain this through the telephone, we can describe the steps as follows:
`log_3(x*(x+2)) = 1`
`log_3(x^2+2x) = 1`
`x^2+2x = 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.
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.
Hence, the solution to the given equation` log_3x + log_3(x+1)` is `x=1` .
Posted by mjripalda on May 29, 2013 at 2:35 AM (Answer #1)
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