If x is an integer and 4^x < 100, what is x?

answer is D as both statements are sufficient to answer the question.

For 4^x < 100 to hold true X has to be less than Four, X< 4.

(1) 4^(x + 1) – 4^(x – 1) > 100 or it can be simplified to 15 * 4^(x-1) > 100

choose a numbers and work out which number for X will satisfy this statement.
(i) choose number 1 answer will be <100
(ii) choose number 2 and answer will be <100
(iii) choose number 3 and answer will be >100
Therefore the the value of X>=3
But for 4^x < 100 to hold true X has to be 3. SUFFICIENT

(2) 4^(x + 1) + 4^x > 100 which can be simplified to 5* 4^x > 100 or to 4^x > 20
choose a numbers and work out which number for X will satisfy this statement.
i) choose number 1 answer will be <100
(ii) choose number 2 and answer will be < 100
(iii) choose number 3 and answer will be >100
Therefore the the value of X>=3
But for 4^x < 100 to hold true X has to be 3. SUFFICIENT

Thanks Bunuel....You saved me from getting confused ..the answer given by tirupatibalaji was wrong....Thanks a lot ..

Bunnel
But the original question is 4^x < 100, what is x?
But not
4^x < 1000, what is x?

Or did i miss something?

Yes, I solved for 4^x < 1,000. Maybe because in this case question becomes a little bit harder.

As actually it's 4^x < 100 then from stem we'll have x<4, so x can be 3, 2, 1, ...

And as both statements give x>2 then x can only be 3. So both statements alone are sufficient.

Answer: D.

Thanks Bunuel.

I thought I was TOTALLY wrong, but this edit makes me feel better!

use common logarithm is fast, simple and very direct:

4^x<100

so 2xlg2<2, lg2=0.301, so x<=3

what is x?

(1) 4^x+1-4^x-1>100

so (15/4)*4^x>100

so 2xlg2>3lg2+1-lg3, lg3=0.477, so x>=3

sufficient

(2) 4^x+1+4^x>100

so 5*4^x>100

4^x>20

so 2xlg2>lg2+1

x>=3

sufficient

D

Can someone explain simplifying the equation by factoring out 4^x and so on. I can't seem to wrap my head around it right now.

When you have addition/subtraction between terms with exponents, all you can do is take common.

Can you simplify \(4^5 + 4^6\)?
Note that \(4^5 = 4*4*4*4*4\)
and \(4^6 = 4*4*4*4*4*4\)

\((4*4*4*4*4) + (4*4*4*4*4*4)\)

You can take five 4s common. You will get

\(4*4*4*4*4 ( 1 + 4)\)
\(= 4^5 * 5\)

So here, it is the same concept.

\(4^x + 4^{x+1}\)
You take \(4^x\) common.
\(4^x ( 1 + 4)\)
\(4^x * 5\)

and

\(4^{x+1} - 4^{x-1}\)
You take \(4^{x-1}\) common because the exponent (x-1) is smaller.
\(4^{x-1} ( 4*4 - 1)\) (because \(4^{x+1}\) has two extra 4s so they are left behind}
\(4^{x-1} * 15\)

Since the questions just mentions that x is an integer, then why can't x be smaller than -3 such as -4, -5 and so on? Because \(4^{-x} = \frac{1}{4^x}\). Am i missing something here?

When you take x as a negative integer, your both the statements A and B would not be satisfied.

Question clearly states that x<4, and both equations prove x=3.

Hence, Answer is D.

Note : try taking the negative values and substitute them in the equations, you will get the equations as incorrect. Hence, Negative values not allowed.

\(4^x < 100\)

x can be { 1 , 2 , 3 }

FROM STATEMENT - I ( INSUFFICIENT )

If x = 1 then 4^(x + 1) – 4^(x – 1) > 100 is

\(4^2 – 4^0 < 100\) = \(15 < 100\) Not true....

If x = 3 then 4^(x + 1) – 4^(x – 1) > 100 is

\(4^4 – 4^3 > 100\) is \(256 - 64\) , ie \(192 > 100\) True


FROM STATEMENT - II ( INSUFFICIENT )

If x = 1 then 4^(x + 1) + 4^x > 100 is

\(4^2 + 4^0 < 100\) = \(17 < 100\) Not true....[/b]

If x = 3 then 4^(x + 1) – 4^(x – 1) > 100 is

\(4^4 + 4^3 > 100\) is \(256 + 64\) , ie \(300> 100\) True

FROM STATEMENT - I & II ( INSUFFICIENT )

Further both the statement I & II are insufficient to find the value of x

Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed, answer will be (E).

Hi All,

We're told that X is an integer and 4^X < 100. We're asked for the value of X. This question can be solved by TESTing VALUES. There's also a great pattern behind the question that is asked. Since you're told that X is an INTEGER and 4^X < 100, that really means that X is less than or equal to 3.

1) 4^(X + 1) – 4^(X – 1) > 100

Fact 1 gives you a real ugly inequality, so instead of trying to do math, use the basic math that you already know and TEST VALUES...

IF....
If X = 2 ---> 64 - 4 = 60 which is NOT greater than 100, so x CANNOT = 2
If X = 3 ---> 256 - 16 = 240, which IS greater than 100.

There's no point in testing smaller integers, since the total won't be > 100 and you're NOT ALLOWED to go bigger than 3 (because of the information in the prompt). Therefore X must equal 3.
Fact 1 is SUFFICIENT

2) 4^(X + 1) + 4^X > 100

Fact 2 provides us with a similar situation, with just a slight change in the math.

IF....
If X = 2 ---> 64 + 16 = 80 which is NOT greater than 100, so X CANNOT = 2 (or anything smaller than 2)
If X = 3 ---> 256 + 64 = 320, which IS greater than 100, so X can be 3.
X can't be anything other than 3.
Fact 2 is SUFFICIENT.

Final Answer:

GMAT assassins aren't born, they're made,
Rich

Can anyone please explain why cant we consider x=-4 or x=-5 in the first equation. this also satisfies the equation 4^x <100 and we get x=+3,-3,-4,-5........ So statement 1 is not sufficient. Please help.

(1) 4^(x + 1) – 4^(x – 1) > 100

x = -4 or -5 does not satisfy stmnt 1.

4^(-4 + 1) - 4^(-4-1)
= 4^(-3) - 4^(-5)
= 1/4^3 - 1/4^5

This is not greater than 100.

Only x = 3 satisfies this as well as 4^x < 100.

Testing values definitely looks like the best way to solve this from the responses above. However, if like me the 'what is x' pushed you into trying to calculate x, was there actually a way of doing so? Conversely, any tips to see through that temptation and go straight to testing values?

We are pretty much solving for x in the solution above: https://gmatclub.com/forum/if-x-is-an-i ... ml#p835476

How do you solve 4^x < 100?
If x is an integer, you figure out the values of x for which this will hold.
Otherwise, you might have to use log tables etc which are not given to you in GMAT.

So when we say x could be 3/2/1/0/-1/-2/-3... (i.e. x <= 3), we have found the values of x for which this works.

Similarly, for statements 1 and 2.

Target question: What is the value of x?

Given: x is an integer and 4^x < 100
Let's take a moment to understand what this tells us about the value of INTEGER x
4^3 = 64, and 4^4 = 256
So, if 4^x < 100, then we know that x ≤ 3

Statement 1: 4^(x + 1) – 4^(x – 1) > 100
Factor the left side to get: 4^(x - 1)[4^2 - 1] > 100
Simplify: 4^(x - 1)[15] > 100
Divide both sides by 15 to get: 4^(x - 1) > 100/15
In other words: 4^(x - 1) > 6.66666...
4^0 is NOT greater than 6.6666....
4^1 is NOT greater than 6.6666....
4^2 IS greater than 6.6666....
4^3 IS greater than 6.6666....
. . . etc.

This tells us (x - 1) ≥ 2
Add 1 to both sides to get: x ≥ 3
We also know that x ≤ 3
We can combine these to write: 3 ≤ x ≤ 3
This means x MUST equal 3
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: 4^(x + 1) + 4^x > 100
Factor the left side to get: (4^x)[4^1 - 1] > 100
Simplify: (4^x)[3] > 100
Divide both sides by 3 to get: 4^x > 100/3
In other words: 4^x > 33.333...
4^1 is NOT greater than 33.333...
4^2 is NOT greater than 33.333...
4^3 IS greater than 33.333...
4^4 IS greater than 33.333...
4^5 IS greater than 33.333...
. . . etc.

This tells us that x ≥ 3
We also know that x ≤ 3
We can combine these to write: 3 ≤ x ≤ 3
This means x MUST equal 3
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Answer: D

Cheers,
Brent

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