Question 92167
Given:
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{{{b(t) = 1.353(1.9025)^t}}}
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In 2004 six years will have passed since the time in years measured from 1998. So you need
to define t as being 6 for this problem. That makes the problem become:
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{{{b(t) = 1.353(1.9025)^6}}}
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Raising 1.9025 to the sixth power can be done directly on a scientific calculator or with
a regular calculator you just multiply:
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1.9025*1.9025*1.9025*1.9025*1.9025*1.9025
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The answer to this part of the problem is that 1.9025 raised to the sixth power is 47.41851975
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Substituting this results in the problem becoming:
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{{{b(t) = 1.353*47.41851975}}}
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and doing the multiplication results in the final answer of:
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{{{b(t) = 64.15725723}}}
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So the answer is that there should have been 64.15725723 million broadband users in 2004.
Rounding this to the nearest million as specified by the problem reduces the answer to 
64 million.
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Hope this helps you to understand the problem.