If 2^x = 3, then which of the following must be true?

Question

If  2^x = 3 , then which of the following must be true?

A.  1 \frac{1}{3} < x < 1 \frac{1}{2}

B.  1 \frac{1}{2} < x < 1 \frac{2}{3}

C.  1 \frac{2}{3} < x < 1 \frac{3}{4}

D.  1 \frac{3}{4} < x < 1 \frac{5}{6}

E.  1 \frac{5}{6} < x < 2

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M37-42

Bunuel · Accepted Answer

Official Solution: If 2^x = 3 , then which of the following must be true? A. 1 \frac{1}{3} < x < 1 \frac{1}{2} B. 1 \frac{1}{2} < x < 1 \frac{2}{3} C. 1 \frac{2}{3} < x < 1 \frac{3}{4} D. 1 \frac{3}{4} < x < 1 \frac{5}{6} E. 1 \frac{5}{6} < x < 1 Square: 2^{2x} = 3^2=9 ; (2^{2x} = 9) > 2^3 , so 2x > 3 ; x>1 \frac{1}{2} Cube: 2^{3x} = 3^3=27 ; (2^{3x} = 27) < 2^5 , so 3x < 5 ; x < 1 \frac{2}{3} . So, 1 \frac{1}{2} < x < 1 \frac{2}{3} . Answer: B

TestPrepUnlimited · Answer

We can do some manipulations to help us determine how big x is in 2^x = 3 . If we square both sides we can get 2^{2x} = 3^2 = 9 . 9 is a little bit bigger than 2^3 . So we have 2^{2x} > 2^3 . It follows that 2x > 3 and x > \frac{3}{2} . We could choose B here since we know x should be just a little bit bigger than \frac{3}{2} , but if we want to be safe we can try to find an upper bound for x. Let's try using the fact that 2^{3x} = 3^3 = 27 . We want an upper bound this time so we have 2^{3x} < 32 = 2^5 and 3x < 5 , which is x < \frac{5}{3} . Hence we can comfortably choose B. Ans: B

yashikaaggarwal · Answer

Answer is supposedly B.
A. 2^4/3= 2.51 < x < 2^3/2= 2.82 (less than 2^x=3)

B. 2^3/2 =2.82 < x < 2^5/3= 3.17 (3 lies in between)

C. 2^5/3= 3.17 < x < 2^7/4= 3.36 (More than 2^x=3)

D. 2^7/4= 3.36 < x < 2^11/6= 3.56 (More than 2^x=3)

E. 2^11/6= 3.56 < x < 2^2= 4 (More than 2^x=3)

B is the answer

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nick1816 · Answer

x = log_2 3 x= \frac{log_{10}3}{log_{10}2} x= \frac{0.47}{0.3} 1.6>\frac{0.47}{0.3}>1.5 OR 2^{1.5} = 2^{(1+0.5)} = 2*2^{0.5} = 2*1.414 = 2.82 2^{1+\frac{2}{3}} = 2*2^{\frac{2}{3}} = 2*4^{\frac{1}{3}} = 2* ^{\frac{1}{3}} ~ 2* ^{\frac{1}{3}} = 2*(\frac{8}{5}) = 3.2 Or (2^3)*(\frac{9}{8}) = 3^2 2^3*1.125 = 3^2 We know that 2^0= 1 and 2^1 = 2 1.125 must be in between 2^{0.1} and 2^{0.2} 2^3*2^{0.1} < 2^{2x} < 2^3*2^{0.2} 2^{3.1} < 2^{2x} < 2^{3.2} 2^{1.55}< 2^x< 2^{1.6} B

AnirudhaS · Answer

I solved this by "feeling". Let me explain. But I need experts to critic me on this method.  nick1816  let me know your thoughts please?

So we know 
2^1.1
2^1.2
2^1.3
.....
2^1.9
2^2

is going to be an exponential curve (NOT linear).

And if it was linear then 2^1.5 would be exactly 3, but its not.
So 2^1.5 < 3
But the rate of growth of this exponential curve is only slight (not as much as say 2^2, 2^3, 2^4...)
So 2^1.5 should be quite close to 3

I mean then I chose the option B which is closest. However, I do understand that this is not full proof (as we do not know how close 2^1.5 is to 3 unless we work it out)

nick1816 · Answer

You are on point bhai. Since the curve is exponential, 2^1.5 < 3.

Now you could find the approximate value of  2^{(1\frac{2}{3})}  just to make sure that you're marking the correct option. I've already mentioned one way to approximate  2^{(1\frac{2}{3})} in my solution.

Another way is-

3375 < 4000 < 4096

15^3<4000<16^3

1.5^3 < 4 < 1.6^3

1.5 < 4^{(\frac{1}{3})} < 1.6

1.5 < 2^{(\frac{2}{3})} < 1.6

1.5*2 < 2*2^{(\frac{2}{3})} < 1.6*2

3 < 2^1*2^{(\frac{2}{3})} < 3.2

3 < 2^{(1+\frac{2}{3})} < 3.2

satya2029 · Answer

2^x=3
2^x-2=1
2^(x-1)=3/2=1.5
 \sqrt{2}=1.414 
0.5<x<0.6
B:)

Bambi2021 · Answer

Must admit I failed this question. But I found some valuable approximations of x. The fractions represent the closest distances between a power of 2 and a power of 3.

https://oeis.org/A254351/internal

"log(3)/log(2) = 1.5849625... is an irrational number. The fractions (2/1, 3/2, 5/3, 8/5, 11/7, 19/12, 46/29, 65/41, 84/53, 317/200, 401/253, 485/306, 569/359, 1054/665, ...) are a sequence of approximations to log(3)/log(2), where each is an improvement on its predecessors."

This means that 2^(19/12) is closer to 3 than 2^(11/7).

"Seven/eleven" seems easy to remember as a decent approximation. =)

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Adit_ · Answer

Well the most logical way I tried is to recursively try getting closer and closer to 3.
Meaning:
Start with 2^1.5 Now since we know root(2)=1.4 ish

2^1.5 = 2.8 ish (very close to 3)
2^2 = 4

between 1.5 and 2 there is a bridging of 1.2 points (4-2.8)
meaning on avg its taking 1.2/5=0.24 (each division from 1.5 to 2 is on avg adding up 0.24 to 2.8)
And 2.8+0.24>3
So we can conclude it has to be between 1.5 and 1.66 only for the powers

Hence Option B

========================================

Is this a right approach? Would love to have your thoughts  Bunuel

If 2^x = 3, then which of the following must be true?

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If 2^x = 3, then which of the following must be true?

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