Algebra II / Mr. Hansen 11/19/2005

Name: _________________________

Solution Key for Test through §6-11

Time limit: 26 minutes (39 minutes for extended time). Calculator is allowed throughout.
Point values: Name = 5 pts., #1 = 25 pts., #2-15 = 5 pts. each. Total = 100 pts.

Part I: Proof (25 points)
1.	Given: a > 0, a ¹ 1, b > 0, b ¹ 1, x > 0 Prove: loga x = (logb x)/(logb a) Proof:
	_______________________________________________________________________
	1. Let y = loga x	| 1. loga x exists since a > 0, a ¹ 1, x > 0
	2. a^y = x	| 2. Def. of log
	3. logb a^y = logb x	| 3. Def. of fcn. (well-defined ans. |     when applied to both sides of eqn.)
	4. y logb a = logb x	| 4. Prop. of logs (F period’s “House Rule”)
	5. y = (logb x)/(logb a)	| 5. Div. prop. of =
	6. loga x = (logb x)/(logb a)	| 6. Trans. prop. (steps 1, 5)
	(Q.E.D.)

Part II: Short Answer (5 points each, little or no partial credit)
All calculator results must be correct to at least 3 decimal places.
2.	3 [but note: questions 4, 5, and 6 should not be computed with a calculator since x is unknown!]
3.	1.956
4.	logx 15 = logx (3 · 5) = logx 3 + logx 5 » .4 + .6 = 1 by LPSL
5.	logx 5^10 = 10 logx 5 » 10 (.6) = 6 by prop. of log of a power (F period’s “House Rule”)
6.	logx (5/3) = logx 5 – logx 3 » .6 – .4 = .2 by LQDL
7.	15 or 16 (either is acceptable) Note: I also accepted “guess and check” answers in that vicinity. However, please note that because so many people gave false answers to #4, #5, and #6 with meaningless extra precision, no calculator will be allowed on the re-test on Monday, Nov. 21. That way, you will need to use what you know instead of relying on the calculator as a crutch.
8.	If you said 15, the reason is found in #4: logx 15 » 1 Þ x^1 » 15. If you said 16, the reason is logx 2 » .25 Þ x^0.25 » 2 Þ (x^0.25)^4 » 2^4 Þ x » 16.
9.	log 347 » 2.540
10.	1.369 · 10^743 Most of the work can be performed on the calculator. Here are the steps: 300^300 = (3 · 10^2)^300             = 3^300 · (10^2)^300             = (3^100)^3· 10^600 » (5.15 . . . · 10^47)^3 · 10^600             = (A · 10^47)^3 · 10^600 (We are letting A represent the mantissa of 5.15 . . . · 10^47.) Calculator hint: To avoid retyping, STO the mantissa of 5.15 . . . · 10^47 into variable A as shown below. 3^100 ENTER Ans/1E47 ENTER [Note: You get the little “E” to appear with the 2nd EE command.] STO®A ENTER Since A^3 » 136.9, we now continue as follows: 300^300 » (A · 10^47)^3 · 10^600             = A^3 · (10^47)^3 · 10^600 » 136.9 · 10^141 · 10^600             = 136.9 · 10^741             = 1.369 · 10^743
11.	Method 1: Take square root of the unrounded answer from #10:
	Now, split up the 10^371.5 factor as (10^0.5)(10^371). If you have been careful to STO your intermediate results, without rounding, you should now have 3.700 · 10^371.
	Method 2: Rewrite problem as (300^300)^0.5 = 300^150 = (3 · 10^2)^150 = 3^150 · (10^2)^150 by PPME, substitution, and distribution. This becomes 3^150 · 10^300 » (3.700 · 10^71) · 10^300, which is again 3.700 · 10^371.
12.	y = 5^x [Note: Many other answers are possible. However, y = x^15 is not correct since that is a power function, not an exponential function. Remember, exponential means that x is in the exponent. A function such as y = 2^3 is also not correct, since that is what we would call a constant function.]
13.	log5 y = x [Note: You received full credit here if your answer was consistent with #12.]
14.	parameter
15.	By far the most common error in both classes was forgetting that the square root of x^2 is |x|. If you doubt this, take your graphing calculator and enter the function Y1=Ö(X^2) followed by ZOOM 6. Do you see that the result is the same as the function y = |x|? Correct answer for E period: 5^15|x|^15c^6/35^3 or approximately 711,780.248|x|^15c^6 Correct answer for F period: 6^15|x|^15c^6/36^3 or 10,077,696|x|^15c^6