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above720
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Exponents & Roots: Data Sufficiency 29 Jan 2007, 22:18
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How would you solve?

8^a(1/4)^b=?

(1) b = 1.5a
(2) a = 2



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Yurik79
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Exponents & Roots: Data Sufficiency 29 Jan 2007, 23:02
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above720
How would you solve?

8^a(1/4)^b=?

(1) b = 1.5a
(2) a = 2
Show more

I am not sure how to read the question is it
8^a*(1/4)^b=
or
8^(a*(1/4)^b)=

try to solve the first one

8^a*(1/4)^b=
{(1/2)^-3a}*{(1/2)^-2b)}
(1)statement b = 1.5a substitute in original equation
{(1/2)^-3a}*{(1/2)^(-2*1.5a)}=
{(1/2)^-3a}*{(1/2)^(-3a)}=8^2a
2)is clearly insuff
Hence my answer is A
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AK
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Exponents & Roots: Data Sufficiency 30 Jan 2007, 05:34
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above720
How would you solve?

8^a(1/4)^b=?

(1) b = 1.5a
(2) a = 2
Show more


if the question is

8^a * (1/4)^b

= 2^3a * (1/2)^ 2b

(1) b = 1.5a

then 2^3a * (1/2)^ 2b = 2^3a * (1/2) ^ 3a =1

Ans = A


if the question is 8^[a(1/4)]^b
then Ans= C
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yezz
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Exponents & Roots: Data Sufficiency Updated on: 30 Jan 2007, 07:31
Originally posted by yezz on 30 Jan 2007, 07:21.
Last edited by yezz on 30 Jan 2007, 07:31, edited 1 time in total.
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How would you solve?

8^a(1/4)^b=?

(1) b = 1.5a
(2) a = 2

8^a(1/4)^b= = 2^3a * 2^-2b = 2^3a -2 b

from one

THE EXPRESSION IS DOWN TO 2^0 = 1......SUFF

from two

no info about value of b.......not suff

ANSWER IS INDEED A , THANKS SUMITHRA
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yezz
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Exponents & Roots: Data Sufficiency 30 Jan 2007, 07:30
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SUMITHRA , THANKS MAN U R RIGHT

I FORGOT THE CRAZY 2 :)

ANSWER IS THUS A
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axl_oz
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Exponents & Roots: Data Sufficiency 30 Jan 2007, 07:31
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8^a(1/4)^b=?

(1) b = 1.5a
(2) a = 2

(8^a)(1/4)^b
Subsitute b from 1

(8)^a(1/4)^1.5a

This an example of (a)^x (b)^x. Cannot be solved. Hence B, C, E

From 2 it is not possible to find a value of the equation. Hence C, E
Combining 1 and 2

You can easily find the value of the equation. Hence C is the best answer
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Sumithra
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Exponents & Roots: Data Sufficiency 30 Jan 2007, 08:53
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yezz
SUMITHRA , THANKS MAN U R RIGHT

I FORGOT THE CRAZY 2 :)

ANSWER IS THUS A
Show more


You are welcome.

Guess, you just need to relax before the big day.

You'll do fine. All the best.



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Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
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