GMATPASSION wrote:

Bunuel wrote:

How many factors does the integer X have?

1. X^(x+3) = (2x)^(x-1)

2. |3x-7| = 2x+2

(1) \(x^{(x+3)}=(2x)^{(x-1)}\) --> \(x^{(x+3)}=2^{(x-1)}*x^{(x-1)}\) --> \(x^{(x+3-x+1)}=2^{(x-1)}\) --> \(x^4=2^{(x-1)}\) --> \(x=1\).

\(x^4=2^{(x-1)}\) --> \(x=1\)---------How do we arrive at this. How to solve these????? just by testing values vaguely or is there any particular mathematics we can do with this?????

I believe Bunuel has already explained this. Just another way to look at it is this:

We end up with: \(x^4=2^{(x-1)} \to x = 2^{\frac{(x-1)}{4}}\) .

We know that if x is an integer, the power of 2,i.e. \(\frac{x-1}{4}\) has to be a non-negative integer.

I.\(\frac{x-1}{4}>0\) As x equals some positive power of 2, x is even.Also, the only way x can be an integer is if the exponent of 2 is raised to an integer\(\to\) x is a multiple of 4.

If (x-1) is a multiple of 4, then x has to be odd and this contradicts out previous statement, that x is even.

II. \(\frac{x-1}{4}=0\)\(\to x=1\) and this satisfies the equality.

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All that is equal and not-Deep Dive In-equality

Hit and Trial for Integral Solutions