() Solve for x: (a) e x = 5 3x SET We take natural log on both sides: ln(e x ) = ln(5 3x ) x = 3 x ln(5) Now we take log base on both sides: log ( x ) = log (3 x ln 5) x = log (3 x ) + log (ln(5)) x x log 3 = log (ln 5) x( log 3) = log (ln 5) x = log (ln 5) log 3 Since the initial equation is defined on all real numbers, plugging this solution in the initial equation makes sense, so this is indeed a solution. (b) log (x + ) + log (x + ) = This equation is defined when the arguments inside the logarithms are positive, that is, when x + > 0 and x + > 0. Thus, the solutions will be in the interval (, + ). We proceed to solve the equation: log (x + )(x + ) = We raise to the power of both sides log (x+)(x+) = (x + )(x + ) = x + 3x + = x + 3x = 0 x(x + 3) = 0 So the possible solutions are x = 0 and x = 3. But since we were looking only at solutions in the interval (, + ), the only solution is x = 0.
() Let f(x) = cos(4x). What is the amplitude? What is the period? Graph the function on the interval [0, 4π] The amplitude is, the period is π = π. In the graph below, note that since the 4 period is π/, we have the portion of the graph from 0 to π/ (π/ is approximately.6) repeating itself 8 times in the interval [0, 4π] (3) Solve for x: ( tan x + π ) = 3 We know that the angle π/6 has tangent 3, and that we obtain all such angles by going around the circle additional multiples of π in both directions. Therefore, x + π = π 6 + kπ, x = π 6 π + kπ, x = π 3π + kπ, 6 x = π 3 + kπ, (4) Draw the appropriate right triangle and evaluate ( ( )) cos arcsin 0 We get the following triangle, where the third side length is given by the Pythagorean Theorem : 0 = 99 = 3. The angle we are interested in is θ, and cos(θ) = 3 0. 0 θ 3
() Solve for x: (a) 3x+3 = 5 x+ SET We take log base on both sides: log ( 3x+3 ) = log (5 x+ ) 3x + 3 = (x + ) log 5 3x x log 5 = log 5 3 x(3 log 5) = log 5 3 x = log 5 3 3 log 5 Since the initial equation is defined on all real numbers, plugging this solution in the initial equation makes sense, so this is indeed a solution. (b) ln(x + 7) ln(x + ) = ln(x 3) This equation is defined when the arguments inside the logarithms are positive, that is, when x + 7 > 0, x + > 0 and x 3 > 0. Thus, the solutions will be in the interval (3, + ). We proceed to solve the equation: ln x + 7 = ln(x 3) x + We raise e to the power of both sides e ln x+7 x+ = e ln(x 3) x + 7 x + = x 3 x + 7 = (x 3)(x + ) x + 7 = x x 3 x 3x 0 = 0 Using the quadratic formula, we have the possible solutions x = and x = 5. But since we were looking only at solutions in the interval (3, + ), the only solution is x = 5.
() Let f(x) = 5 sin ( ) 3x. What is the amplitude? What is the period? Graph the function on the interval [0, 4π] The amplitude is 5, the period is π 3 = 4π. In the graph below, note 3 that since the period is 4π/3, we have the portion of the graph from 0 to 4π/3 (4π/3 is approximately 4.) repeating itself 3 times in the interval [0, 4π] (3) Solve for x: cos(x) = We know two angles that have cosine : π 4 that have cosine π and. We obtain half the angles 4 by going around the circle multiples of π from π/4, and we obtain the other half of angles by going around the circle additional multiples of π from π/4. Therefore, we have two kinds of angles: π 4 + kπ, and π 4 + kπ, (4) Draw the appropriate right triangle and evaluate tan ( arccos ( 7 We get the following triangle, where the third side length is given by the Pythagorean Theorem : 7 = 45 = 3 5. The angle we are interested in is θ, and tan(θ) = 3 5. )) 7 3 5 θ
() Solve for x: (a) 3x = 4 x SET 3 We rewrite 3x = 4 x as 3x = ( ) x, so 3x = x. Now we take log base from both sides, and log ( 3x ) = log ( x ) 3 x = x 3 x = x+ Now we take log base again on both sides, and we get log (3 x ) = log ( x+ ) x log 3 = x + x log 3 x = x(log 3 ) = x = log 3 Since the initial equation is defined on all real numbers, plugging this solution in the initial equation makes sense, so this is indeed a solution. (b) log 0 (x + 3) + log 0 (x + 4) = log 0 (6) This equation is defined when the arguments inside the logarithms are positive, that is, when x + 3 > 0 and x + 4 > 0. Thus, the solutions will be in the interval ( 3, + ). We proceed to solve the equation: log 0 (x + 3)(x + 4) = log 0 (6) We raise 0 to the power of both sides 0 log 0 (x+3)(x+4) = 0 log 0 (6) (x + 3)(x + 4) = 6 x + 7x + = 6 x + 7x + 6 = 0 Using the quadratic formula, we get the possible solutions to be x = and x = 6. But since we were looking only at solutions in the interval ( 3, + ), the only solution is x =.
() Let f(x) = 3 sin ( ) 4x 3. What is the amplitude? function on the interval [0, 3π] What is the period? Graph the The amplitude is 3, the period is π 4 = 3π. In the graph below, note that since the 3 period is 3π/, we have the portion of the graph from 0 to 3π/ (3π/ is approximately 4.7) repeating itself times in the interval [0, 4π] (3) Solve for x: sin(x) = We know two angles that have sine : π and π + π. We obtain half the angles 6 6 that have sine by going around the circle multiples of π from π, and we obtain 6 the other half of angles by going around the circle additional multiples of π from π + π. Therefore, we have two kinds of angles: 6 and π 6 + kπ, π + π 6 + kπ, (4) Draw the appropriate right triangle and evaluate sin (arctan(5)) We get the following triangle, where the third side length is given by the Pythagorean Theorem : + 5 = 6. The angle we are interested in is θ, and sin(θ) = 5 6. 6 5 θ