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sunniboy007
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Sunni 3/3/04 Q4 03 Mar 2004, 17:04
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Solve for 'x' if 25^x-2/sqrt5 = (1/5)^x-7.5

5
-3
4
-5
:x

ps. I know we can plug and play here, but please show your reasoning. Cheers!



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Paul
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Sunni 3/3/04 Q4 03 Mar 2004, 20:00
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sunniboy007
Solve for 'x' if 25^x-2/sqrt5 = (1/5)^x-7.5

5
-3
4
-5
:x

ps. I know we can plug and play here, but please show your reasoning. Cheers!


Please put parentheses and brackets to distinguish between exponents and order of operations. Question is not clear as it is, especially the left hand side
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Sunni 3/3/04 Q4 03 Mar 2004, 20:04
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Sorry about that.

Solve for 'x' if
25^(x-2)/(sqrt5) = (1/5)^(x-7.5)

Hope that is clearer
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Paul
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Sunni 3/3/04 Q4 03 Mar 2004, 20:08
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sunniboy007
Sorry about that.

Solve for 'x' if
25^(x-2)/(sqrt5) = (1/5)^(x-7.5)

Hope that is clearer


Is it 25^(x-2) which is divided by sqrt5 or is it only ^(x-2) which is divided by sqrt5?
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Sunni 3/3/04 Q4 03 Mar 2004, 20:11
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base and the exponential divided by sqrt5
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Paul
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Sunni 3/3/04 Q4 03 Mar 2004, 20:23
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Let's start with the left hand side
[25^(x-2)] / sqrt5
= [5^2(x-2) / 5^1/2]
= 5^(2x-4-1/2)
= 5^(2x-9/2)
Now let's deal with right hand side
(1/5)^(x-7.5)
= 5^(-1(x-15/2))
= 5^(-x+15/2)
Now let's set up the inequality with the exponents since we have the same base
2x - 9/2 = -x + 15/2
3x = 24/2
3x = 12
x = 4
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Sunni 3/3/04 Q4 04 Mar 2004, 06:52
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I got 5.

25^(x-2)/(sqrt5) = (1/5)^(x-7.5)

25^(x-2)/5^1/2 = (5^-1)^x-15/2
(25/5)^x-2-1/2 = 5^-x+15/2
5^x-5/2 = 5^15/2-x
x - 5/2 = 15/2 - x
2x = 20/2
x = 5
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Sunni 3/3/04 Q4 04 Mar 2004, 06:57
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Hmmm. I see where I made my error. Instead of putting 25 in exponential form, I divided it by 5.


Paul is right. Ans = 4.



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