would the answer for 1/x + 3/2x be 7/2x or 5/2x or netheir and how would u get it because i got 7/2x and my friend got 5/2x...wat is the correct answer and how do u get it?? thnxx
Assuming the 2x is all in the denominator in the second term it is 5/(2x). To solve this you must first get a common denominator. Between x and 2x the common denominator is 2x. To change the first term, multiply it by 2/2
The answer is 5/2x because you have to look at it like two fractions. These two fractions are (1/x) and (3/2x). When you're finding the sum of fractions (or the difference for that matter) you have to find the lowest common denominater for both of the fractions. In this case, it's 2x, so you have to multiply the first by (2/2), which makes it (2/2x) and then the sum of 2/2x and 3/2x is 5/2x.
Addition of two logarithms may well be rewritten as multiplication like so: log[ (3x - a million) * (2x+3) ] = log (4x) log( 6x^2 + 7x - 3 ) = log(4x) The words interior the logarithms might desire to be equivalent if the logarithms themselves are equivalent, so: 6x^2 + 7x - 3 = 4x 6x^2 + 3x - 3 = 0 3(x^2 + x - a million) = 0 x^2 + x - a million = 0 x = (-a million +/- sqrt(a million+4))/2 [quadratic formulation, or finished the sq., whichever] x = (-a million +sqrt(5))/2 [We dismiss the unfavorable answer, on the grounds that could make 4x unfavorable, and you are able to no longer take the logarithm of unfavorable numbers. desire that enables!
Comments
Assuming the 2x is all in the denominator in the second term it is 5/(2x). To solve this you must first get a common denominator. Between x and 2x the common denominator is 2x. To change the first term, multiply it by 2/2
1/x + 3/(2x)
2/(2x) + 3/(2x)
5/(2x)
The answer is 5/2x because you have to look at it like two fractions. These two fractions are (1/x) and (3/2x). When you're finding the sum of fractions (or the difference for that matter) you have to find the lowest common denominater for both of the fractions. In this case, it's 2x, so you have to multiply the first by (2/2), which makes it (2/2x) and then the sum of 2/2x and 3/2x is 5/2x.
Answer: 5/2x
Addition of two logarithms may well be rewritten as multiplication like so: log[ (3x - a million) * (2x+3) ] = log (4x) log( 6x^2 + 7x - 3 ) = log(4x) The words interior the logarithms might desire to be equivalent if the logarithms themselves are equivalent, so: 6x^2 + 7x - 3 = 4x 6x^2 + 3x - 3 = 0 3(x^2 + x - a million) = 0 x^2 + x - a million = 0 x = (-a million +/- sqrt(a million+4))/2 [quadratic formulation, or finished the sq., whichever] x = (-a million +sqrt(5))/2 [We dismiss the unfavorable answer, on the grounds that could make 4x unfavorable, and you are able to no longer take the logarithm of unfavorable numbers. desire that enables!
1/x + 3/(2 x)
=
5
--------
2 x