Inequality 2

1. This doesn't help much. This statement is equivalent to 120 < 4^x - 4^(x -1). So you are looking for the difference between two consecutive powers of 4. Just picture this: 4^4 = 256, while 4^3 = 64. The difference between the two is obviously greater than 120 and this gets bigger with bigger powers.

2. x^2 = 36 has two solutions:
a.x = -6
b. x = 6
We cannot tell from 2 alone what is the value of x.

Now, if we put both statements together, what do we get?
Well, let's see:
a. For x = -6, x -1 = -7. 4^(-6) = 1/(4^6) is smth greater than 0 but smaller than 1 and so is 4^(-7) = 1/(4^7). So the difference between these two numbers is obviously not greater than 120.
b. x = 6 is consistent with all the constraints, since 4^6 - 4^5 will be greater than 120.

Answer C.

where am I going wrong

2. x^2 = 36
x is either -6 or 6
3x will then either be -18 or 18, which are both less than 500

I agree with Dana if the question was asking for the value of x, but the question is asking is 3x <500

Should answer be not B

sacx wrote:where am I going wrong

2. x^2 = 36
x is either -6 or 6
3x will then either be -18 or 18, which are both less than 500

I agree with Dana if the question was asking for the value of x, but the question is asking is 3x <500

Should answer be not B

:roll: IMO B

And I agree with sacx's explanation.

ooops.... Sorry, I was in a hurry... Yes guys you're right...

This is how I tackled it (assuming it's about 3^x, not 3x):

From the question:
3^x < 500
implies
x < 6

So we're checking to see if we know that x< 6.


From 1:
4^(x-1) + 120 < 4^x
120 < 3 * 4^(x-1)
40 < 4^(x-1)
x > 2
Insufficient


From 2:
x^2 = 36
x = 6 or x = -6
Insufficient


Combining 1 and 2:
Only x = 6 satisfies both constraints. So it is definitely not the case that x < 6.

Answer C.