Algebra II / Graham, Hansen, James

Answer Key to Feb. 2000 Sample Test on Ch. 6

1.(a)

3
(As stated in the note above, the answer by itself is worth zero points. You must show your work, since getting the answer alone on the calculator is almost trivial. The same is true for nearly every problem on this test.)
Work: Let x be the answer. Since x = log_1/6 (1/216), we know (1/6)^x = 1/216. Thus 1^x/6^x = 1/216 Þ 1/6^x = 1/216 Þ 6^x = 216 Þ x = 3. Reason: We learned that 216 is the cube of 6 (math fact).

1.(b)

748 (no credit for answer without work)
Work: 10^{log 17 + log 44} = 10^{log 17} 10^{log 44} = 17 (44) = 748

2.(a)

log₄ 686 (work required)

2.(b)

log₅₀₀₅ 1, or 0 if simplified (work required)
Work: By LPSL and LQDL, log₅₀₀₅ 7 + log₅₀₀₅ 143 – log₅₀₀₅ 5005 + log₅₀₀₅ 5 = log₅₀₀₅ (7 · 143 · 5 / 5005 ) = log₅₀₀₅ 1 = 0.

3.(a)

S = {17} (work required)
Work: We know x^–2 = 1/289 Þ 1/x² = 1/289 Þ x² = 289 Þ x = ± 17. We discard the negative solution since the base of a logarithm must be positive.

3.(b)

S = {–1.5 + 0.25Ö108} (approximately 1.098)
Note: There is another solution (approx. –4.098) that must be rejected since we cannot compute the log of a negative number.

3.(c)

S = {4 log₁₅ 9.5}

4.(a)

–35.9799 (work required)
Work: loggg 150 = loggg (2 · 3 · 5²) = loggg 2 + loggg 3 + 2 loggg 5 = –4.9773 + (–7.8888) + 2 (–11.5569) = –35.9799.

4.(b)

For typical bases such as 10 (or actually any base greater than 1), logs are always positive for numbers such as 2, 3, and 5—since the exponent you would apply to such a base would have to be positive in order to obtain values such as 2, 3, and 5. However, for bases less than 1, you need a negative exponent (i.e., log) to obtain values such as 2, 3, and 5. (Try it; positive exponents just make your answer shrink.) Therefore loggg has a base less than 1. Since bases must always be positive, we conclude that the base of loggg is between 0 and 1.

4.(c)

Let b denote the base of loggg. Since log_b 3 = –7.8888, the definition of logs tells us b^–7.8888 = 3, so log b^–7.8888 = log 3 Þ –7.8888 log b = log 3 Þ log b = log 3 / (–7.8888) = –0.06048. Now, if log b = –0.06048, b itself must be 10^–0.06048, or 0.8700 (answer).

5.

8 (work needed)
Work: Using change of base, we get –3 log₃ (1/3) + 2 (log₃ 243 / log₃ 9) = –3 log₃ (3^–1) + 2 (log₃ 3⁵ / log₃ 3²) = –3 (–1) + 2 (5/2) = 3 + 5 = 8.

6.(a)

f ^–1(x) = (x² – 4)/2 (work required)
Note: In precalculus, you will learn that technically this is not correct. The domain of f ^–1 must be restricted to match the range of f, namely the interval [0, ¥ ). Thus to be totally correct, we should say f ^–1(x) = (x² – 4)/2 where x Î [0, ¥ ). For tomorrow’s test it would be OK to say f ^–1(x) = (x² – 4)/2.

6.(b)

f ^–1(f (x)) = f ^–1((2x + 4)^1/2) = {[(2x + 4)^1/2]² – 4}/2 = (2x + 4 – 4) / 2 = 2x / 2 = x

6.(c)

It is assumed that you can use your calculator to do this. Do you notice how f and f ^–1 form a mirror image about the line y = x? (To see the mirror-image effect fully, you have to restrict the domain of f ^–1 to [0, ¥ ) as described in part (a).)

7.

Let M denote morale in mors. Then M = ab^t (general equation). Using algebra (preferred) or ExpReg (allowed in Mr. Hansen’s class), you can get a = 150, b = 0.99823, from which you can write the particular equation M = 150 · 0.99823^t. Plugging in 265 for t gives M = 93.8 mors, so yes, the morale level does remain high all year long. (Believe me, 90 mors is pretty good.)

8.

S/N ratio (in dB) = 10 log₁₀ (45/0.018) = 10 log 2500 = 10 (3.39794) = 34 dB. By the way, that’s not very good; S/N ratios of 50 or 60 dB are desirable. Your stereo would sound quite lousy with a 34 dB S/N ratio.

9.

10^1880.742 (acceptable answer)
5.5212 ´ 10¹⁸⁸⁰ (preferred format)