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Find the value of x if 256^(x - 2) = 2^(-x - 7).

 Jan 6, 2020
 #1
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+1

Solve for x :
256^(x - 2) = 2^(-x - 7)

-256^(x - 2) = (-1)^1·2^(8 x - 16):
(-1)^1·2^(-x - 7) = (-1)^1·2^(8 x - 16)

Equate exponents of -1 and 2 on both sides:
1 = 1 and -x - 7 = 8 x - 16

All equations give x = 1 as the solution:
 
x = 1

 Jan 6, 2020
 #2
avatar+27 
+3

I'm not sure where you learned this, but there is a lot of extraneous math in your solution. All the "1^ -1" is completely unnecessary and confusing, although it does somehow give the correct solution. Here's a simpler way to do it:


256x-2 = 2-x-7 | 256 is equal to 28, so you can make the bases the same

28x-16=2-x-7 | Now, you can take away the bases and solve algebraically. If you perform log2 on both sides, you're left with just the exponents, so I'm just not showing a step.

8x-16=-x-7 | Next, you solve algebraically.

8x+x-16+16=-x+x-7+16

9x=9 | Divide both sides by nine.

x=1 | You have your answer!

 

Now, let's plug it back in to check if it works.

 

256x-2 = 2-x-7 | Replace x with our answer, 1

2561-2=2-1-7 | Change 256 to 28

28-16=2-1-7 | Simplify

2-8=2-8

 

It works! It may look hard at first, but once you try it out it's much easier and saves you the extra steps.

 

TLDR; Match your bases, take the log and solve algebraically.

 Jan 6, 2020
 #3
avatar+23497 
+3

2^8 is 256 , so this is equivalent to:

28(x-2) =2(-x-7)    Now equate the exponents

 

8(x-2) = -x-7

8x-16 = -x-7

9x = 9

x=1

 Jan 6, 2020

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