\[ \text { ClufN } y=\log _{g} 8 \] (a) \( \log _{q} 64 \) (b) \( \log _{2} 2 \) (c) \( \log _{q} \frac{16}{q} \)

We will show one acceptable way to express the answers in terms of y and the base‐conversion factor logg q. (Any answer equivalent to the ones below is correct.) Suppose you are given   y = logg 8      [that is, g^y = 8]. Then we want to find:   (a) logq 64   (b) log₂ 2   (c) logq (16/q). It is a good idea to use the change‐of‐base formula and properties of logarithms. ––––––––––––––––––––––––––––––– Step 1. Express some useful numbers in “g–terms” Notice that   8 = 2³  so that logg 8 = logg (2³) = 3·logg 2. But we are given logg 8 = y, so   logg 2 = y/3. ––––––––––––––––––––––––––––––– Step 2. Solve each part. (a) Evaluate logq 64. Write 64 as a power. Since 64 = 8² (because 8² = 64), we have   logq 64 = logq (8²) = 2·logq 8. Now use the change‐of–base formula to convert logq 8 into logarithms with base g:   logq 8 = (logg 8)/(logg q). But logg 8 = y, so   logq 8 = y/(logg q). Thus   logq 64 = 2·(y/(logg q)) = (2y)/(logg q). ––––––––––––––––––––––––––––––– (b) Evaluate log₂ 2. This one is immediate. For any positive base a (with a ≠ 1) we have   loga a = 1. So   log₂ 2 = 1. ––––––––––––––––––––––––––––––– (c) Evaluate logq (16/q). Recall that   logq (16/q) = logq 16 – logq q      [by the quotient rule]. Also, logq q = 1. Furthermore, write 16 as a power of 2:   16 = 2⁴, so that   logq 16 = logq (2⁴) = 4·logq 2. Now, use the change‐of‐base formula:   logq 2 = (logg 2)/(logg q). But we already found logg 2 = y/3, so   logq 2 = (y/3)/(logg q) = y/(3·logg q). Thus,   logq 16 = 4·[y/(3·logg q)] = (4y)/(3·logg q). Finally,   logq(16/q) = (4y)/(3·logg q) – 1. ––––––––––––––––––––––––––––––– Final Answers (a) logq 64 = (2y)/(logg q). (b) log₂ 2 = 1. (c) logq (16/q) = (4y)/(3·logg q) – 1. Any answer equivalent to the ones above is correct.