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Nookie1986 [14]
3 years ago
5

Solve for x

2%20%20%7B%7D%5E%7B%20%5Cfrac%7B1%7D%7B2%7D%20%7D%20" id="TexFormula1" title="8 {}^{2 + x} + 2 {}^{3x} = 32 {}^{ \frac{1}{2} } " alt="8 {}^{2 + x} + 2 {}^{3x} = 32 {}^{ \frac{1}{2} } " align="absmiddle" class="latex-formula">
​
Mathematics
1 answer:
Stella [2.4K]3 years ago
7 0

Answer:

8^2+x + 2^ 3x = 32^ 1/2

2^3(2+x) + 2^3x = 2^5^1/2

All two's will cancel out

3(2+x) + 3x = 5^1/2

6+3x + 3x = 5^1/2

6 +6x = 2.23

6x = 2.23 - 6

6x = -3.77

x = -3.77÷ 6

x = -0.63

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Please answer my question
vladimir1956 [14]

I will refer to the graphs as 1 - 6, starting at the top and going down. 6 is not present, though I will still include it by process of elimination.

All real numbers less than -7 or odd numbers between -7 and 7, exclusive. = Graph #6

All real numbers except -7. = Graph #2

Numbers divisible by 4 between -10 and 10. = Graph #4

All real numbers up to and including -1 or more than 1. = Graph #1

All real numbers between -7 and -4, inclusive, or between -1 and 1, exclusive. = Graph #3

Prime numbers between -10 and 10. = Graph #5

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7 0
3 years ago
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yKpoI14uk [10]
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Answer:

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