Question

practice: change bases
if ( a^{\frac{b}{2}} = 25 ) for positive integers ( a ) and ( b ), what is one possible value of ( b )?

answer box

check
explain

practice: raise to a negative exponent
if ( n^{-\frac{1}{3}} = x ), where ( n > 0 ), what is ( n ) in terms of ( x )?

choose 1 answer:

a ( -\frac{1}{x^3} )

b ( \frac{1}{x^3} )

c ( -sqrt3{x} )

d ( sqrt3{x} )

check

Explanation:

First Sub - Question (Finding a possible value of \(b\))

Step 1: Recall the exponent rule for fractional exponents

The expression \(a^{\frac{b}{2}}\) can be rewritten using the rule \(x^{\frac{m}{n}}=\sqrt[n]{x^{m}}\) (or equivalently \((\sqrt[n]{x})^{m}\)). So, \(a^{\frac{b}{2}}=\sqrt{a^{b}}\) (or \((\sqrt{a})^{b}\)). We know that \(a^{\frac{b}{2}} = 25\). Let's consider some positive integer values for \(a\) and \(b\).

Step 2: Try simple values for \(a\)

Let's start with \(a = 5\). Then we have \(5^{\frac{b}{2}}=25\). But \(25 = 5^{2}\), so if we set the exponents equal (since the bases are the same and positive), we have \(\frac{b}{2}=2\).

Step 3: Solve for \(b\)

To solve \(\frac{b}{2}=2\) for \(b\), we multiply both sides of the equation by 2. So, \(b=2\times2 = 4\). We can also check with other values of \(a\), for example, if \(a = 25\), then \(25^{\frac{b}{2}}=25\), so \(\frac{b}{2}=1\) and \(b = 2\). But we just need one possible value, and 4 (or 2) is a valid solution.

Step 1: Recall the negative exponent rule

The negative exponent rule states that \(x^{-m}=\frac{1}{x^{m}}\) (for \(x
eq0\) and \(m\) a real number). So, \(n^{-\frac{1}{3}}=\frac{1}{n^{\frac{1}{3}}}\), and we know that \(n^{-\frac{1}{3}}=x\), so \(\frac{1}{n^{\frac{1}{3}}}=x\).

Step 2: Recall the fractional exponent rule for roots

The expression \(n^{\frac{1}{3}}\) is equivalent to \(\sqrt[3]{n}\) (by the rule \(x^{\frac{1}{n}}=\sqrt[n]{x}\) for \(x\geq0\) and \(n\) a positive integer). So we can rewrite \(\frac{1}{\sqrt[3]{n}}=x\).

Step 3: Solve for \(n\)

First, take the reciprocal of both sides of the equation \(\frac{1}{\sqrt[3]{n}}=x\). We get \(\sqrt[3]{n}=\frac{1}{x}\). Then, to eliminate the cube root, we cube both sides of the equation. Cubing the left - hand side, \((\sqrt[3]{n})^{3}=n\), and cubing the right - hand side, \((\frac{1}{x})^{3}=\frac{1}{x^{3}}\). So, \(n=\frac{1}{x^{3}}\).

Answer:

4 (or 2, or other valid positive integers. For example, if \(a = 625\), then \(625^{\frac{b}{2}}=25\), \(625 = 25^{2}\), so \(25^{2\times\frac{b}{2}}=25^{1}\), \(25^{b}=25^{1}\), \(b = 1\) is also possible. But 4 is a common and simple solution)

Second Sub - Question (Expressing \(n\) in terms of \(x\))