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Chapter 2: Problem 10

The numbers are too large to be handled by a calculator. These exercises require an understanding of the concepts. Write \(2^{3000}\) as a power of 8 .

Short Answer

Expert verified
The expression \(2^{3000}\) can be written as a power of 8 as \(8^{1000}\).

Step by step solution

01

Express 8 as a power of 2

First, we need to express the number 8 as a power of 2. Since \(2^3 = 8\), we have 8 as a power of 2.
02

Express \(2^{3000}\) in terms of 8

Now, to represent \(2^{3000}\) as a power of 8, we need to rewrite it in terms of the base 8, using the fact that \(8 = 2^3\).
03

Substitute the power of 2

To do this, we replace 8 with \(2^3\): \(2^{3000} = (2^3)^n\),
04

Calculate the new exponent

Now, we need to find the exponent n. Using exponent properties, we can simplify the right-hand side of the equation: \((2^3)^n = 2^{3n}\) So, we have: \(2^{3000} = 2^{3n}\) Since the bases are equal (both are 2), the exponents must also be equal: \(3000 = 3n\) Divide both sides by 3: \(n = 1000\)
05

Rewrite the expression as a power of 8

Now that we have found the exponent n, we can rewrite the expression \(2^{3000}\) as a power of 8: \(2^{3000} = (2^3)^{1000} = 8^{1000}\) The expression \(2^{3000}\) can be written as a power of 8 as \(8^{1000}\).

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Most popular questions from this chapter

Write each expression as the sum of a polynomial and a rational function whose numerator has smaller degree than its denominator. $$ \frac{2 x+1}{x-3} $$

Suppose \(p(x)=3 x^{7}-5 x^{3}+7 x-2\) (a) Show that if \(m\) is a zero of \(p\), then $$ \frac{2}{m}=3 m^{6}-5 m^{2}+7 $$ (b) Show that the only possible integer zeros of \(p\) are \(-2,-1,1,\) and 2 . (c) Show that no integer is a zero of \(p\).

Suppose \(s(x)=\frac{x^{2}+2}{2 x-1}\) (a) Show that the point (1,3) is on the graph of \(s\). (b) Show that the slope of a line containing (1,3) and a point on the graph of \(s\) very close to (1,3) is approximately -4 [Hint: Use the result of Exercise \(25 .]\)

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Suppose \(q\) is a polynomial of degree 4 such that $$ \begin{array}{r} q(0)=-1 . \text { Define } p \text { by } \\ \qquad p(x)=x^{5}+q(x) . \end{array} $$ Explain why \(p\) has a zero on the interval \((0, \infty)\).
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