M37-26

Question

If  3^{(2x+1)} + 4(3^x) – 20 = 0 , then what is approximate value of  x  ?

A.  \frac{1}{6} 
B.  \frac{1}{5} 
C.  \frac{1}{3} 
D.  \frac{2}{3} 
E.  \frac{9}{10}

Bunuel · Answer

Official Solution:

If  3^{(2x+1)} + 4(3^x) – 20 = 0 , then what is approximate value of  x  ?

A.  \frac{1}{6} 
B.  \frac{1}{5} 
C.  \frac{1}{3} 
D.  \frac{2}{3} 
E.  \frac{9}{10}

3^{(2x+1)} + 4(3^x) – 20 = 0 ;
 
 3*3^{(2x)} + 4(3^x) – 20 = 0 ;
 
 3*(3^x)^2 + 4(3^x) – 20 = 0 ;
 
Denote  3^x  as  k :  3k^2 + 4k– 20 = 0 ;
 
Solve for  k :  k=2  or  k=-\frac{10}{3}  (discard this root because  3^x  is positive and cannot equal to a negative number).
 
We get  3^x=2 
 
Now,  3^2 \approx 2^3 , so  3^{\frac{2}{3}} \approx 2 , thus  x\approx {\frac{2}{3}} .

Answer: D

phmahi1997 · Answer

I like the solution - it’s helpful.

chamhi · Answer

How can you assume that 3^x is positive? We do not whether x is an even or odd number.

Bunuel · Answer

3^x is always positive, regardless of whether x is odd, even, positive, negative, or even a fraction. It never becomes zero or negative. I suggest brushing up on fundamentals of number properties and exponents.

chamhi · Answer

Clearly did not have enough sleep yesterday. I misread it for x^3 (in this case, whether x was positive or negative would have had impact)...

Rebaz · Answer

Bunuel

I understand your solution until 3^x = 2. But what you do from here on, i cannot follow your reasoning behind 3^2 ≈ 2^3, so 3^2/3 ≈ 2, thus X ≈ 2/3.

Can you please shed more light on it?

Thanks in advance!

Bunuel · Answer

We are given that 3^x = 2.

Now, observe that 3^2 ≈ 2^3, so we can write:

3^(2/3) ≈ (3^2)^(1/3) ≈ (2^3)^(1/3) = 2

Thus, since 3^x = 2, and 3^(2/3) ≈ 2, it follows that x ≈ 2/3.

danzoliv · Answer

3k^2+4k–20=0
how can I easily find the roots for k?

Many thanks for considering my request.

Bunuel · Answer

Check the links below: Factoring Quadratics: https://www.purplemath.com/modules/factquad.htm Solving Quadratic Equations: https://www.purplemath.com/modules/solvquad.htm Hope it helps.

shashag · Answer

Great question, but I wonder why it's put as a medium-level question in the forum discussion haha. definitely seems tougher than that, especially under exam conditions.

komatsu · Answer

I like the solution - it’s helpful.

findingmyself · Answer

I did not quite understand the solution. What is wrong in solving this question like:

consider X=1, then 27+12=20 not possible

So X<1
What about X=1/2

9+4 root 3=20
Too low

So X is some where betwee 1/2 and 1
Two options: 2/3 or 9/10

How will we filter down from here? Is there a way to reach solution from here.

Bunuel · Answer

You can find alternative solutions here: https://gmatclub.com/forum/gmat-club-ol ... 67171.html

M37-26

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