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stich3 [128]
3 years ago
6

If 1/64=4^2s-1 times 16^2s+2 what is the value of s?

Mathematics
2 answers:
Irina18 [472]3 years ago
5 0

Answer: s=-1

Explanation:

We can re-write the equation as follows:

\frac{1}{64}=(4^{2s-1})(16^{2s+2})

Let's rewrite the terms on the left and on the right as follows:

2^{-6} = ((2^2)^{2s-1})((2^4)^{2s+2})

So we have

2^{-6} = (2^{2(2s-1)})(2^{4(2s+2)})

Now we can use the following rule:

x^{a+b}=x^a x^b

to rewrite the term on the right:

(2^{2(2s-1)})(2^{4(2s+2)})=2^{2(2s-1)+4(2s+2)}

So now we have

2^{-6}=2^{2(2s-1)+4(2s+2)}

and since the bases are the same, we can just equalize the exponents:

-6=2(2s-1)+4(2s+2)

-6=4s-2+8s+8\\-12 = 12 s\\s=-1

galina1969 [7]3 years ago
4 0
S=-1 see attachment for explanation

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