SAT SUBJECT TEST MATH LEVEL 2
PART 2
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REVIEW OF MAJOR TOPICS
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CHAPTER 1
Functions
1.4 Exponential and Logarithmic Functions
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The basic properties of exponents and logarithms and the fact that the exponential function and the logarithmic function are inverses lead to many interesting problems.
The basic exponential properties:
For all positive real numbers x and y, and all real numbers a and b:

The basic logarithmic properties:
For all positive real numbers a, b, p, and q, and all real numbers x, where a
1 and b
1:

The basic property that relates the exponential and logarithmic functions is:
For all real numbers x, and all positive real numbers b and N, logb N = x is equivalent to bx = N. If the base is the number e, ln, the natural logarithm, is used instead of loge.
By convention, the base is 10 if no base is indicated.
|
TIP logbN is only defined for positive N. |
EXAMPLES
1. Simplify xn–1 · x2n · (x2–n)2
This is equal to xn–1 · x2n · x4–2n = xn–1 + 2n + 4–2n = xn+3.
2. Simplify
.
In order to combine exponents using the properties above, the base of each factor must be the same.

3. If log 23 = z, what does log 2300 equal?
log 2300 = log(23 · 100) = log 23 + log 100 = z + log 102 = z + 2
Note: Examples 3 and 4 can be easily evaluated with a calculator.
4. If ln 2 = x and ln 3 = y, find the value of ln 18 in terms of x and y.
ln 18 = ln(32 · 2) = ln 2 + 2ln 3 = x + 2y
5. Solve for x: log b(x + 5) = logb x + logb 5.
logb x + logb 5 = logb (5x)
Therefore, log(x + 5) = log(5x), which is true only when:

6. Evaluate
.

The last equality implies that
27x= 3
(33)x= 3
33x= 31
Therefore, 3x = 1 and x =
.
Thus,
.
You could also use the change-of-base formula and your calculator.

Therefore,
.
The graphs of all exponential functions y = bx have roughly the same shape and pass through point (0,1). If b > 1, the graph increases as x increases and approaches the x-axis as an asymptote as x decreases. The amount of curvature becomes greater as the value of b is made greater. If 0 < b< 1, the graph increases as x decreases and approaches the x-axis as an asymptote as x increases. The amount of curvature becomes greater as the value of b is made closer to zero.

The graphs of all logarithmic functions y = logb x have roughly the same shape and pass through point (1,0). If b > 1, the graph increases as x increases and approaches the y-axis as an asymptote as x approaches zero. The amount of curvature becomes greater as the value of b is made greater. If 0 < b < 1, the graph decreases as x increases and approaches the y-axis as an asymptote as x approaches zero. The amount of curvature becomes greater as the value of b is made closer to zero.

EXERCISES
1. If xa· (xa+1)a· (xa)1–a = xk, then k =
(A) 2a + 1
(B) a + a2
(C) 3a
(D) 3a + 1
(E) a3 + a
2. If log8 3 = x · log2 3, then x =
(A) ![]()
(B) 3
(C) 4
(D) log4 3
(E) log8 9
3. If log10 m =
, then log10 10m2 =
(A) 2
(B) 2.5
(C) 3
(D) 10.25
(E) 100
4. If logb 5 = a, logb 2.5 = c, and 5x= 2.5, then x =
(A) ac
(B) ![]()
(C) a + c
(D) c – a
(E) The value of x cannot be determined from the information given.
5. If f (x) = log2 x, then ![]()
(A) ![]()
(B) 1
(C) ![]()
(D) ![]()
(E) 0
6. If ln (xy) < 0, which of the following must be true?
(A) xy < 0
(B) xy < 1
(C) xy > 1
(D) xy > 0
(E) none of the above
7. log2 m =
and log7 n =
, mn =
(A) 1
(B) 2
(C) 96
(D) 98
(E) 103
8. Log7 5 =
(A) 1.2
(B) 1.1
(C) 0.9
(D) 0.8
(E) – 0.7
9.
=
(A) 1.9
(B) 2.0
(C) 2.1
(D) 2.3
(E) 2.5
10. If $300 is invested at 3%, compounded continuously, how long (to the nearest year) will it take for the money to double? (If P is the amount invested, the formula for the amount, A, that is available after t years is A = Pe0.03t .)
(A) 26
(B) 25
(C) 24
(D) 23
(E) 22
Answers and Explanations
Exponential and Logarithmic Functions
1. (C) ![]()
2. (A)
3. (A) log(10m2) = log 10 + 2 log m = 1 + 2 ·
= 2.
4. (B) ba = 5, bc = 2.5 = 5x , using the relationships between logs and exponents: (ba )x = bax = 5x = bc . Therefore, ax = c and
.
5. (B)
6. (B) Since ln stands for loge, and e > 1, xy < 1.
* 7. (D) Converting the log expressions to exponential expressions gives
and
. Therefore, mn =
.
* 8. (D)
.
* 9. (C) ![]()
* 10. (D) Substitute in A = Pe0.03t to get 600 = 300e0.03t. Simplify to get 2 = e0.03t . Then take ln of both sides to get ln 2 = 0.03t and t =
. Use your calculator to find that t is approximately 23.
