﻿ Exponential and Logarithmic Functions - Functions - REVIEW OF MAJOR TOPICS - SAT SUBJECT TEST MATH LEVEL 2 ﻿

## PART 2 ## REVIEW OF MAJOR TOPICS ## CHAPTER 1Functions

### 1.4 Exponential and Logarithmic Functions 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 and y, and all real numbers and b: The basic logarithmic properties:

For all positive real numbers a, b, p, and q, and all real numbers x, where 1 and 1: The basic property that relates the exponential and logarithmic functions is:

For all real numbers x, and all positive real numbers and N, logb N 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 = + log 102 = + 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(+ 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 > 1, the graph increases as increases and approaches the x-axis as an asymptote as decreases. The amount of curvature becomes greater as the value of is made greater. If 0 < b< 1, the graph increases as decreases and approaches the x-axis as an asymptote as increases. The amount of curvature becomes greater as the value of is made closer to zero. The graphs of all logarithmic functions = logb x have roughly the same shape and pass through point (1,0). If > 1, the graph increases as increases and approaches the y-axis as an asymptote as approaches zero. The amount of curvature becomes greater as the value of is made greater. If 0 < < 1, the graph decreases as increases and approaches the y-axis as an asymptote as approaches zero. The amount of curvature becomes greater as the value of is made closer to zero. EXERCISES

1.       If xa· (xa+1)a· (xa)1–a = xk, then =

(A)  2+ 1

(B)  a + a2

(C)  3a

(D)  3+ 1

(E)  a3 + a

2.       If log8 3 = · log2 3, then =

(A) (B)  3

(C)  4

(D)  log4 3

(E)  log8 9

3.       If log10 , 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 =

(A)  ac

(B) (C)  a + c

(D)  c – a

(E)  The value of 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.       logm = and logn = 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 is the amount invested, the formula for the amount, A, that is available after years is 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 )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 Pe0.03t to get 600 = 300e0.03t. Simplify to get 2 = e0.03t . Then take ln of both sides to get ln 2 = 0.03and . Use your calculator to find that is approximately 23.

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