devinawilliam83 wrote:
if x^2 >x^3 : we write x^2-x^3 >0 ie we subtract and dont divide
in this case we are dividinge is that right?
If the variable is given to be non zero, you can divide in an equation and if you also know the sign of the variable, you can divide in an inequality too. (and adjust the inequality according to the sign of the variable)
e.g. \(x^3 = x\). What values can x take?
x can be 0 so we don't divide.
\(x^3 - x = 0\)
\(x(x^2 - 1) = 0\)
\(x = 0, 1, -1\)
Here, if you forget about 0 and divide by x, you get x^2 = 1 so x = +1 or -1. You lose out one solution: x = 0
But in case you have
\(x^3 = x\). What values can x take if x is non zero?
Now you can divide and you will get x = 1 or -1
Similarly, say you have an inequality: \(x^2 > x^3\)
If you divide by x^2 thinking it's non negative, you get the range x < 1 which is incorrect since you have to specify that 'x cannot be 0'. It holds for every value less than 1 other than 0.
\(x^2 > x^3\) does not hold for x = 0.
So instead, you can choose to solve in two ways:
1. Take cases and use division: \(x^2 > x^3\)
If \(x \neq 0\), we can divide. We get x < 1.
If x = 0, the inequality does not hold. Hence, \(x \neq 0\)
Answer: x < 1, \(x \neq 0\)
2. Subtract: x^3 - x^2 < 0
x^2 (x - 1) < 0
One of the terms must be negative and the other positive. Since x^2 cannot be negative, x - 1 must be negative (so x < 1) and x^2 must be positive so \(x \neq 0\)
Answer: x < 1, \(x \neq 0\)
Either way, you will get the same answer (obviously!)
In this question, as Bunuel said, \(9^q\) must be positive. It cannot take value 0 or negative for any value of q so there are no complications at all. You can easily divide.
One little thing Karishma.
then it's clear that \(x\neq{0}\) thus \(x^2>0\) and you can safely reduce by it and write \(x>1\).