For symbolic solutions to exponential or logarithmic equations, choose Solve + Exact.
Enter Variable(s) to Solve for if requested.
For numerical solutions, you can either enter a coef cient in decimal notation and choose Solve + Exact or, apply Evaluate Numerically to the symbolic solutions. In the case of a single variable, you can choose Solve + Numeric.
I Solve + Exact
3x= 8, Solution is: ln 31 (2i k + ln 3 log38) j k 2 Z
ex= y + 1
y 1, (Solve for x) Solution is:
( n ; if y = 1
lny11(y + 1) + 2i k j k 2 Zo
if y 6= 1 ,
For simpli ed results, you may want to use the options Principal Value Only and Ignore Special Cases.
I To obtain solutions in the following form
1. Choose Tools + Engine Setup and select the General page.
2. Check both Principal Value Only and Ignore Special Cases.
I Solve + Exact
3x= 8, Solution is: log38 ex= y + 1
y 1, (Solve for x), Solution is: ln 1
y 1(y + 1) P = Qekt, (Solve for k), Solution is: 1tlnPQ
For numerical solutions, you can either enter a coef cient in decimal notation and choose Solve + Exact or, apply Evaluate Numerically to the symbolic solutions. For a particular solution, you can choose Solve + Numeric.
I Solve + Exact
3x= 8:0, Solution is: 1: 892 8
log (3x + y) = 8:0, Solution is: 993: 65 0:333 33y
I Solve + Numeric
3x= 8, Solution is: f[x = 1: 892 8]g
log (3x + y) = 8, Solution is: f[x = 842: 07; y = 454: 74]g
Exercises
1. Given that when x2 3x + 5k is divided by x + 4 the remainder is 9, nd the value of k using Divide on the Polynomials submenu and Solve + Exact.
2. De ne functions f(x) = x3+ x ln x and g(x) = x + ex. Evaluate f(g(x)), g(f(x)), f (x)g(x), and f (x) + g(x).
3. Find the equation of the line passing through the two points (x1; y1), (x2; y2).
4. Find the equation of the line passing through the two points (2; 5), (3; 7).
5. Find the equation of the line passing through the two points (1; 2), (2; 4).
6. Find the slope of the line determined by the equation sx + ty = c.
7. Find the center and semi-axes of the ellipse 16 x2+ 4y2+ 96x 8y + 84 = 0.
8. Factor the difference of powers xn ynfor several values of n, and deduce a general formula.
9. Applying Factor to x2+ p
5 3 x 3p
5 gives the factorization x2+ p
5 3 x 3p
5 = x +p
5 (x 3)
showing that the system can factor some polynomials with irrational roots. However,
applying Factor to x2 3 and x3+ 3x2 5x + 1 does not do anything. Find a way to factor these polynomials.
10. Find all real and complex solutions to the system of equations 2x2 y = 1
x + 3y3 = 4
Solutions
1. Using Polynomials + Divide, x2 3x + 5k ax + by + c = 0 through these two points. Substituting these points in the equation for the line gives the two equations ax1+ by1+ c = 0 and ax2+ by2+ c = 0.
Applying Solve + Exact to the system
ax1+ by1+ c = 0 ax2+ by2+ c = 0
gives many possibilities, depending on the relationship between the points (x1; y1) and (x2; y2). For a short solution that works for most pairs, chose Tools + Engine Setup and check Ignore Special Cases. This gives the solution
a = cy1 cy2 x1y2 x2y1
; b = cx1 cx2 x2y1 x1y2
Consequently, the equation for the line is c y1 y2
x1y2 x2y1
x c x1 x2
x2y1 x1y2
y + c = 0 or, clearing fractions and collecting coef cients by factoring in place,
(y1 y2) x (x1 x2) y + (x1y2 y1x2) = 0 4. For the points (2; 5), (3; 7), the system of equations is
2a + 5b + c = 0 3a 7b + c = 0
Apply Solve + Exact to obtain
Solution is: a = 12
29c; b = 1 29c Consequently, the equation for the line is
12
29cx 1
29 cy + c = 0 or, clearing fractions and simplifying,
12x + y + 29 = 0
5. Since the point (0; 0) lies on the line, you do not get a unique solution to the system of equations for the pair a; b. Thus, choosing Solve + Exact and specifying a; b for the variables gives no response. However, specifying a; c for Variable(s) to Solve for gives the solution
[a = 2b; c = 0]
Thus, the equation for the line is
2bx + by = 0 or, dividing by b and applying Simplify,
( 2bx + by)1
b = 2x + y = 0
Note An interesting method for nding the equation of a line through two speci ed points using determinants is described in a Matrix Algebra exercise on page 368.
6. The slope-intercept form of the equation for a line is y = mx + b, where m is the slope and b the y-intercept. If a line is given as a linear equation in the form sx + ty = c, you can nd the slope by solving the equation for y. Apply Expand to the solution y = sx ct to get y = ct stx, revealing the slope to be st.
7. To nd the center and semi-axes of the ellipse 16x2+ 4y2+ 96x 8y + 84 = 0, a. Subtract 84 from both sides of the equation to get
16 x2+ 4y2+ 96x 8y + 84 84 = 0 84
b. Select the left side and while holding down the CTRLkey, apply Simplify; then, do the same to the right side to obtain
96x 8y + 16x2+ 4y2= 84
c. Drag the terms containing x together, select them and click ; then, do the same to the terms containing y, to get
16x2+ 96x + 4y2 8y = 84
d. Factor out the leading coef cients by dragging them outside the parentheses and dividing other coef cients by their value, to get
16 x2+96
16x + 4 y2 8
4y = 84
e. Add the product of the coef cient of x2with the square of one-half the coef cient
of x to both sides; then, do the same for y, to get
16 x2+9616x + 129616 2 + 4 y2 84y + 1284 2
= 84 + 16 129616 2+ 4 1284 2
f. Select the term 16 x2+9616x + 129616 2 and while holding down theCTRLkey, apply Factor; then, do the same for the term with y, to get
16 (x + 3)2+ 4 (y 1)2= 84 + 16 1
g. Select the right side and while holding down theCTRLkey, apply Simplify, to get 16 (x + 3)2+ 4 (y 1)2= 64
h. Divide each term by the right-hand side, to get 16 (x + 3)2
64 +4 (y 1)2
64 = 64
i. Select each term and while holding down theCTRL64key, apply Factor, to get 1
4(x + 3)2+ 1
16(y 1)2= 1
You can read the answer from this form of the equation: The center of the ellipse is ( 3; 1), and the semi-axes arep
4 = 2 andp 16 = 4.
8. Apply Factor to several differences.
x2 y2= (x y) (x + y)
After looking at only these few examples, you might nd it reasonable to conjecture that, for n odd,
xn yn = (x y)
n 1X
k=0
xn k 1yk We leave the general conjecture for you. Experiment.
9. Using the clue from the example that the system will factor over roots that appear as coef cients, apply Factor to the product p
3 x2 3 to get p
3 x2 3 = p3 x p
3 x +p
3 . Now you can divide out the extraneousp 3 to get
5 to factor this polynomial:
p5 x3+ 3x2 5x + 1 = p
5 (x 1) x p
5 + 2 x +p
5 + 2 . Then, can-celing the extraneous factor ofp
5, you have x3+ 3x2 5x + 1 = (x 1) x + 2 +p
5 x + 2 p
5
10. With the insertion point in the array
2x2 y = 1 x + 3y3 = 4 choose Solve + Exact. You receive the response
Solution is : [y = 1; x = 1] ; x = 4 3y3; y = 1 where 1is a root of 53Z^ 53Z^2+ ^Z3+ ^Z4+ ^Z5 3118 Leave your insertion point in the polynomial
5 3Z^ 5
3Z^2+ ^Z3+ ^Z4+ ^Z5 31
and choose Polynomials + Roots. You receive the following solution.18
roots: vector of roots, and click the parentheses button. Type an x at the left of the vector, leave the insertion point in the expression, and apply Evaluate for the following:
x
To display this result, you can concatenate the two vectors: Place them side by side and from the Matrices submenu, choose Concatenate. Then, select the (two-column) matrix, choose Edit + Insert Row(s) to add two new rows at the top. Label the columns with x and y and, in the other new row, add the solution x = 1, y = 1.
x= 3y3+4 y
1 1
1:0466 1:1893
0:60247 + 0:40783i 0:60663 0:98268i 0:60247 0:40783i 0:60663 + 0:98268i 0:62562 0:36793i 0:48801 0:92069i 0:62562 + 0:36793i 0:48801 + 0:92069i
4 Trigonometry
Trigonometry developed from the study of triangles, particularly right triangles, and the relations between the lengths of their sides and the sizes of their angles. The trigonomet-ric functions that measure the relationships between the sides of similar triangles have far-reaching applications that extend well beyond their use in the study of triangles.