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by sudhir3127 » Sat Aug 02, 2008 8:24 am
My answer is 5/2

If 4^4x = 1600, what is the value of (4x–1)^2

i think u wrongly posted the question as (4x-1)^2 rather it shud be
4^(x-1)^2

solve this

4^2x-2 which is 4^2x/16 --------------------------------1

Now solve this square rooting on both the sides
4^4x = 1600

we will be left with ..

4^2x = 40-------------------------------------- 2


using 1 and 2 we get

40/16 = 5/2,,,

Hope it helps..
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by raunekk » Sat Aug 02, 2008 8:39 am
exactly...!! i guess thr is sum typing error..

i have been tryin tis for last 20 mins..:)
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by dbart06 » Sat Aug 02, 2008 8:45 am
trying to follow the math:
how did you you go from 4^2x-2 => 4^2x/16

i am not good with exponents.

Thanks for the input
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Yes Sudhir u r right

by anksbhandari » Sat Aug 02, 2008 8:46 am
YES Sudir u r right

i wrongly posted the question as (4x-1)^2 rather it shud be
4^(x-1)^2

Thanks for your explanation
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by dbart06 » Sat Aug 02, 2008 8:59 am
sudhir3127.

I am not following your math:

4^4x=1600 => 4^2x=40 you took square root of 4x & 1600..i follow..not following how you got 5/2

not following 4^x-1^2..if you mult x-1 * x-1 how is it = to 2x-1.. i get 1

please advise
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by sudhir3127 » Sat Aug 02, 2008 9:08 am
4^x-1^2..if you mult x-1 * x-1 how is it = to 2x-1.. i get 1

u need to know the formula for it ..
a^m^2 is not a^m*m its a^2m.....

in the same way ...

4^x-1^2 is 4^2x-2 which is 4^2x/4^2 ( formula again.. a^m-n= a^m/a^n)..

hope its clear now...
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by dbart06 » Sat Aug 02, 2008 9:15 am
where can I refresh my math..

I don't follow that x-1^2 = 2(x-1)... i thought that when you square a # ie (4) you get 4*4 not 2*4
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by raunekk » Sat Aug 02, 2008 9:42 am
4^4x = 1600

taking square root both the sides

4^2x = 40....(a)

Now, 4^(x-1)^2= 4^2(x-1) = 4 ^ 2x-2 ( a^b^c= a^bc)

Thus, 4^2x/ 4^2

From (a)

40/16

=5/2.

Thus D
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by dbart06 » Sat Aug 02, 2008 9:57 am
still do not follow the math.. i just don't see it:

x-1^2 is 2 (x-1) HOW? this is squared...x-1 * x-1 same as 4 * 4

HELP?
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by malolakrupa » Sat Aug 02, 2008 6:21 pm
(x^2)^3 can be written as x^6 . Hence by the same rule 4^(x-1)^2
can be written as 4^(2x-2) . Applicable only to exponents.
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by gabriel » Sat Aug 02, 2008 9:10 pm
sudhir3127 wrote:
u need to know the formula for it ..
a^m^2 is not a^m*m its a^2m.....


hope its clear now...
Actually no, a^m^2 is not a^2m, (a^m)^2 is a^2m Eg. 2^2^3=2^8 and (2^2)^3=2^6. So if the question is 4^(x-1)^2 then you guys have the wrong solution.
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by gabriel » Sat Aug 02, 2008 9:12 pm
dbart06 wrote:still do not follow the math.. i just don't see it:

x-1^2 is 2 (x-1) HOW? this is squared...x-1 * x-1 same as 4 * 4

HELP?
You are right, (x-1)^2 is not 2(x-1), I suspect the question is still wrong and if it is not then the solution is wrong.
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by rey.fernandez » Sun Aug 03, 2008 12:03 am
As this is a MG problem, I can confirm how the actual problem should read:

If 4^(4x) = 1600, what is the value of [4^(x-1)]^2?

Let's start with the question:
[4^(x-1)]^2 is equivalent to 4^[2(x-1)] or 4^(2x-2).

Remember that (a^b)^c = a^(bc). In other words, when you raise a power to a power, you can rewrite by keeping the same base and multiplying the exponents. Of course, this works in the other direction too, something we'll use later on in the solution.

We can manipulate 4^(2x-2) further as 4^(2x)/4^2 or 4^(2x)/16.

So, to find this value, we need to find a value for 4^(2x).

4^(4x) = 1600
[4^2x]^2 = 1600 (Same exponent property that we used earlier)
4^2x = 40

Now that we have that value, we can fin 4^(2x)/16.

40/16 = 5/2

The correct answer is D.
Rey Fernandez
Instructor
Manhattan GMAT
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by dbart06 » Mon Aug 04, 2008 6:10 am
thanks for the clarity.
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