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exponents

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Source: — Problem Solving |
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by gm800 » Tue May 08, 2007 12:48 pm
For questions like this, I think it makes sense not to solve for x.

from here we have:

4^4x = 1600 (Equation 1)

Now lets take a ratio of what we have vs. whats required:

4^4x / [4^(x–1)]^2 = 4 ^ (2x + 2) (Ratio 1)

From Equation 1, 4^2x = 40
Thus 4^(2x+2) = 40 * 16 = 640

So what we need now is 1600/640 (from Ratio 1)
This simplifies to 5/2

is that correct?
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by ns88 » Tue May 08, 2007 3:33 pm
let's work with 1600 first

1600= 4^2 * 100

1600=4^2 * 4*25

1600=4^3 * 25

since 25 is just above 16 we can say 4^3 * 4^2.1=1600

so, 4^5.1 =1600


now back to the orig: 4^4x=1600

x= 5.1/4


pug x into 2nd equation

[4^(x–1)]^2

(4^1/4)^2

4^1/4 is the same as sqrt root of 2=

ans =2

since we approximate the answer is 2 something, D, 5/2
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by Tame the CAT » Tue May 08, 2007 6:56 pm
I got 5/2
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by Cybermusings » Wed May 09, 2007 4:50 am
Suddenly I stumbled upon an explanation of my own...here goes..
4^4x = 1600; we have to find [4^(x-1)]^2
[4^(x-1)]^2 = 4^(2x-2) [multiplication law of exponents]
Now if we square this we get --- [4^(2x-2)]^2
= 4^(4x-4)
= 4^4x/4^4 (Again using law of exponents)
= Now we know that 4^4x = 1600
Hence 4^4x/4^4 = 1600/16*16 = 25/4
Now [4^(2x-2)]^2 = 25/4
Hence sqr rt of this or [4^(2x-2)] = sqr rt of (25/4) = 5/2
Yippie...I solved it too!
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Re: exponents

by gviren » Wed May 23, 2007 6:16 am
We have to calculate

{4^(x-)}^2 = 4^2(x-1) = 16^(x-1)

Now 4^4x=1600
16^2x=1600
taking sqrt
16^x=40
16^x-1=40/16 = 5/2
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