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# The positive value of x that satisfies the equation (1 + 2x)

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Manager
Joined: 29 Nov 2011
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The positive value of x that satisfies the equation (1 + 2x)  [#permalink]

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16 Apr 2012, 23:54
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26% (02:23) correct 74% (02:15) wrong based on 512 sessions

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The positive value of x that satisfies the equation (1 + 2x)^5 = (1 + 3x)^4 is between

A. 0 and 0.5
B. 0.5 and 1
C. 1 and 1.5
D. 1.5 and 2
E. 2 and 2.5

Bunuel, can you please explain this one?
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Posts: 57030
Re: The positive value of x that satisfies the equation (1 + 2x)  [#permalink]

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17 Apr 2012, 02:13
6
6
Smita04 wrote:
The positive value of x that satisfies the equation (1 + 2x)^5 = (1 + 3x)^4 is between

A. 0 and 0.5
B. 0.5 and 1
C. 1 and 1.5
D. 1.5 and 2
E. 2 and 2.5

Bunuel, can you please explain this one?

Trial and error would probably be the easiest way to solve this problem. When x is large enough positive number, then because of the exponents (5>4), LHS will be more than RHS (as you increase the positive value of x the distance between the values of LHS and RHS will increase).

Try x=1 --> LHS=3^5=81*3=243 and RHS=4^4=64*4=256, so (1 + 2x)^5 < (1 + 3x)^4. As you can see LHS is still slightly less than than RHS. So, the value of x for which (1 + 2x)^5 = (1 + 3x)^4 is slightly more than 1.

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Re: The positive value of x that satisfies the equation (1 + 2x)  [#permalink]

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25 Jun 2013, 12:25
5
1
The positive value of x that satisfies the equation (1 + 2x)^5 = (1 + 3x)^4 is between

A. 0 and 0.5
B. 0.5 and 1
C. 1 and 1.5
D. 1.5 and 2
E. 2 and 2.5

Assume any variable X = LHS - RHS
Calculating values of X at 0, 1 and 2. The answer lies between the two value where the polarity of X changes.
X at 0 = 0
X at 1 = -13
X at 2 = 724

Value lies between 1 and 2
X at 1.5 =109 (approx.) --- polarity changes after value at 1 but remains same for value at 2
=> X lies between 1 and 1.5

Hence answer is C
##### General Discussion
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Joined: 13 Mar 2012
Posts: 207
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17 Apr 2012, 00:28
2
Smita04 wrote:
The positive value of x that satisfies the equation (1 + 2x)^5 = (1 + 3x)^4 is between
(A) 0 and 0.5
(B) 0.5 and 1
(C) 1 and 1.5
(D) 1.5 and 2
(E) 2 and 2.5

Bunuel, can you please explain this one?

i will go with A as per graph they would be meeting at x=0;y=1

hence 0 must be there .
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17 Apr 2012, 02:03
1
@ kraizada84 : which figure is this ?
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17 Apr 2012, 02:31
1
Smita04 wrote:
The positive value of x that satisfies the equation (1 + 2x)^5 = (1 + 3x)^4 is between
(A) 0 and 0.5
(B) 0.5 and 1
(C) 1 and 1.5
(D) 1.5 and 2
(E) 2 and 2.5

Bunuel, can you please explain this one?

i will go with A as per graph they would be meeting at x=0;y=1

hence 0 must be there .

I did a mistake here 0 is not positive hence its wrong.

By the graph of x^4 and x^5 between 1 and 2 they meet again in the 1st quadrant. rest we need to verify the value of each function at 1 and 0.5

Bunuel gave a nice way. I tried to solve graphically.
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18 Apr 2012, 12:22
Smita04 wrote:
The positive value of x that satisfies the equation (1 + 2x)^5 = (1 + 3x)^4 is between
(A) 0 and 0.5
(B) 0.5 and 1
(C) 1 and 1.5
(D) 1.5 and 2
(E) 2 and 2.5

Bunuel, can you please explain this one?

i will go with A as per graph they would be meeting at x=0;y=1

hence 0 must be there .

The question is asking for the positive value of X ..
if i get basics correct ( which i wish to get an expert opinion ) "0" is even integer but
"0" cannot be positive or negative.
hence we cannot go for A.
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18 Apr 2012, 12:34
kashishh wrote:
Smita04 wrote:
The positive value of x that satisfies the equation (1 + 2x)^5 = (1 + 3x)^4 is between
(A) 0 and 0.5
(B) 0.5 and 1
(C) 1 and 1.5
(D) 1.5 and 2
(E) 2 and 2.5

Bunuel, can you please explain this one?

i will go with A as per graph they would be meeting at x=0;y=1

hence 0 must be there .

The question is asking for the positive value of X ..
if i get basics correct ( which i wish to get an expert opinion ) "0" is even integer but
"0" cannot be positive or negative.
hence we cannot go for A.

Zero is an even number and it's neither positive nor negative - TRUE.

But that's not the reason to discard option A (notice that A covers some positive range the same way as other options).
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Re: The positive value of x that satisfies the equation (1 + 2x)  [#permalink]

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25 Jun 2013, 05:55
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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Re: The positive value of x that satisfies the equation (1 + 2x)  [#permalink]

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08 Jul 2013, 18:02
Zuch wrote:
The positive value of x that satisfies the equation (1 + 2x)^5 = (1 + 3x)^4 is between

A. 0 and 0.5
B. 0.5 and 1
C. 1 and 1.5
D. 1.5 and 2
E. 2 and 2.5

Assume any variable X = LHS - RHS
Calculating values of X at 0, 1 and 2. The answer lies between the two value where the polarity of X changes.
X at 0 = 0
X at 1 = -13
X at 2 = 724

Value lies between 1 and 2
X at 1.5 =109 (approx.) --- polarity changes after value at 1 but remains same for value at 2
=> X lies between 1 and 1.5

Hence answer is C

Hi Zuch,
how much time did it take for you to do this?
I guess on exam we might not get options as close as C and D.
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Re: The positive value of x that satisfies the equation (1 + 2x)  [#permalink]

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08 Jul 2013, 23:15
cumulonimbus wrote:
Zuch wrote:
The positive value of x that satisfies the equation (1 + 2x)^5 = (1 + 3x)^4 is between

A. 0 and 0.5
B. 0.5 and 1
C. 1 and 1.5
D. 1.5 and 2
E. 2 and 2.5

Assume any variable X = LHS - RHS
Calculating values of X at 0, 1 and 2. The answer lies between the two value where the polarity of X changes.
X at 0 = 0
X at 1 = -13
X at 2 = 724

Value lies between 1 and 2
X at 1.5 =109 (approx.) --- polarity changes after value at 1 but remains same for value at 2
=> X lies between 1 and 1.5

Hence answer is C

Hi Zuch,
how much time did it take for you to do this?
I guess on exam we might not get options as close as C and D.

Hi Cumulonimbus,

I understand your concern regarding time. Calculating values for 1.5 especially is tedious. I did not time my attempt but I guess it would take approx. 3 minutes.
But, the point here is for this question, I found this method to be more efficient i.e. time-wise. I think it would be one of the difficult ones on GMAT and hence some extra time dedicated to such problems can be justified. Anyways if such questions come, you have to attempt them unless you are short on time and want to guess and move on.
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Re: The positive value of x that satisfies the equation (1 + 2x)  [#permalink]

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26 Feb 2014, 23:11
Its taking too long for me to solve it, any shorter methods please... ?
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Re: The positive value of x that satisfies the equation (1 + 2x)  [#permalink]

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05 May 2015, 02:59
Zuch wrote:
The positive value of x that satisfies the equation (1 + 2x)^5 = (1 + 3x)^4 is between

A. 0 and 0.5
B. 0.5 and 1
C. 1 and 1.5
D. 1.5 and 2
E. 2 and 2.5

Assume any variable X = LHS - RHS
Calculating values of X at 0, 1 and 2. The answer lies between the two value where the polarity of X changes.
X at 0 = 0
X at 1 = -13
X at 2 = 724

Value lies between 1 and 2
X at 1.5 =109 (approx.) --- polarity changes after value at 1 but remains same for value at 2
=> X lies between 1 and 1.5

Hence answer is C

Excellent ad easy explanation...kudos to you Zuch!

Thank you.
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Posts: 15
Re: The positive value of x that satisfies the equation (1 + 2x)  [#permalink]

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28 Aug 2016, 08:34
Zuch wrote:
The positive value of x that satisfies the equation (1 + 2x)^5 = (1 + 3x)^4 is between

A. 0 and 0.5
B. 0.5 and 1
C. 1 and 1.5
D. 1.5 and 2
E. 2 and 2.5

Assume any variable X = LHS - RHS
Calculating values of X at 0, 1 and 2. The answer lies between the two value where the polarity of X changes.
X at 0 = 0
X at 1 = -13
X at 2 = 724

Value lies between 1 and 2
X at 1.5 =109 (approx.) --- polarity changes after value at 1 but remains same for value at 2
=> X lies between 1 and 1.5

Hence answer is C

Hi Zuch,

Can you help me understand the concept that why the answer will lie between the values in which the polarity of 'X' changes?
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The positive value of x that satisfies the equation (1 + 2x)  [#permalink]

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22 Mar 2017, 19:37
1
Smita04 wrote:
The positive value of x that satisfies the equation $$(1 + 2x)^5 = (1 + 3x)^4$$ is between

A. 0 and 0.5
B. 0.5 and 1
C. 1 and 1.5
D. 1.5 and 2
E. 2 and 2.5

OFFICIAL SOLUTION

The two expressions in question, $$(1 + 2x)^5$$ and $$(1 + 3x)^4$$, could in theory be expanded, but you’ll wind up doing a lot of algebra, only to find an equation involving $$x^5$$, $$x^4$$, $$x^3$$, $$x^2$$, and x. Solving for the positive value of x that makes the equation true is nearly impossible.

So how on Earth can you answer the question? Notice that the question does not require you to find a precise value of x; you just need a range. So plug in good benchmarks and track the value of each side of the equation.

Start with x = 1. The “equation” becomes $$3^5 =?= 4^4$$.

Compute the two sides: $$3^5 = 243$$, while $$4^4 = 16^2 = 256$$. So the right side is bigger.

Now, we need another benchmark. Try x = 2. The “equation” becomes $$5^5 =?= 7^4$$.

Compute or estimate the two sides. $$5^4 = 25^2 = 625$$, and multiplying in another 5 gives you something larger than 3,000 (3,125, to be precise). Meanwhile, $$7^4 = 49^2 < 50^2 = 2,500$$, so the left side is now bigger. Somewhere between x=1 and x=2, then, the equation must be true.

The only benchmark left to try is 1.5, or $$\frac{3}{2}$$. Plugging in, you get $$4^5 =?= (\frac{11}{2})^4$$

$$4^5 = 2^{10} = 1,024$$ (it’s good to know your powers of 2 this high)

For the other side, first compute the denominator: 24 = 16. Now the numerator: $$11^4 = 121×11×11 = 1,331×11 = 14,641$$ (also quick to do longhand)

So $$(\frac{11}{2})^4 = \frac{14,641}{16} < 1,000$$. The right side is bigger.

Let’s recap:

$$(1 + 2x)^5 = (1 + 3x)^4$$ at what value of x?

$$(1 + 2x)^5 < (1 + 3x)^4$$ when x = 1

$$(1 + 2x)^5 > (1 + 3x)^4$$ when x = 1.5

$$(1 + 2x)^5 > (1 + 3x)^4$$ when x = 2

So the value at which the two sides are equal must be between 1 and 1.5.

The correct answer is C.

Extra points:

As x grows past 2, the larger power (5) on the left takes over, so you can see that the left side will always be bigger.

What about when x is between 0 and 1? Well, first notice that at x=0, the two sides are again equal. If x is a tiny positive number (say, 0.001 or something), then you can ignore higher powers of x ($$x^2$$, $$x^3$$, etc.), and this simplifies the algebraic expansion:

For tiny positive x,

$$(1 + 2x)^5$$ = 1 + 5(2x) + higher powers of x ? 1 + 10x

$$(1 + 3x)^4$$ = 1 + 4(3x) + higher powers of x ? 1 + 12x

So right away, the right side is bigger than the left side. You can also check x = ½:

$$(1 + 2x)^5 = 2^5 = 32$$

$$(1 + 3x)^4 = (\frac{5}{2})^4 = 2.5^4 = 6.25^2 > 36 (= 6^2) > 32$$

Again, the right side is bigger than the left side. For every x between 0 and 1, in fact, the right side is larger than the left side. This fact isn’t particularly easy to prove, but you don’t need to do so.
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Re: The positive value of x that satisfies the equation (1 + 2x)  [#permalink]

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07 Apr 2018, 11:10
Hi All,

While the question 'looks' a bit crazy, the wording and set-up for it implies that there's just 1 solution (and we won't have to actually calculate the exact value). Second, the answers are numbers, so we should TEST THE ANSWERS. Since the "math" in this question could get ugly, I'm going to stick to integers.

X = 1 is a great place to start. Plug that value into both calculations and you'll end up with...

3^5 and 4^4
243 and 256

So the numbers are real close; the answer has to be close to 1. The question is do we pick B or C? Notice that the SECOND value in the above calculation is BIGGER.

For the next test, you can decide between a few options (.5, 1.5 or 2 if you prefer using an integer).

X = 2 gives us...

5^5 and 7^4
3025 and (49)(49) = approx. (50)(50) = 2500

Notice now that the FIRST value is BIGGER.

This tells me that increasing the value of X makes the first part of the equation "grow" faster than the second part. By extension, at some point between X=1 and X=2, the values would have been equal and then a little past that point the first value became bigger than the second. Thus, I would have to choose an answer that was a little bigger than 1.

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Re: The positive value of x that satisfies the equation (1 + 2x)  [#permalink]

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Re: The positive value of x that satisfies the equation (1 + 2x)   [#permalink] 11 Apr 2019, 12:00
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