Question

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Answer 1

Explanation:

e^(2x + 1) = 3e^(4 - 3x)

e^(2x + 1) = e^3(4- 3x)

lne^ (2x + 1) =ln e^3(4- 3x)

2x + 1 = 3*(4 - 3x)

2x + 1 = 12 - 9x

11x + 1 = 12

11x = 12 - 1

11x = 11

x = 1

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I've never worked with binary logs but I'll give this my best shot. If someone comes along and gives you a different answer, use it.

lg(y-6) + lg(y + 15) = 2

lg (y - 6)(y + 15) = 2

inverse (y - 6)(y + 15) = 2^2

(y - 6)(y + 15) = 4

y^2 + 9y - 90 - 4 = 0

y^2 + 9y - 94 = 0

y = 6.19

y = -15,19

Answer 2

9514 1404 393

Answer:

(a) x = (3 -ln(3))/5 ≈ 0.819722457734

(b) y = 10

Explanation:

(a) Taking the natural log of both sides, we have ...

2x +1 = ln(3) +4 -3x

5x = ln(3) +3 . . . . . . . . add 3x-1

x = (ln(3) +3)/5 ≈ 0.820

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(b) Assuming "lg" means "log", the logarithm to base 10, we have ...

log(y -6) +log(y +15) = 2

(y -6)(y +15) = 100 . . . . . . . taking antilogs

y^2 -9x -190 = 0 . . . . . . . . eliminate parentheses, subtract 100

(y -19)(y +10) = 0 . . . . . . . . factor

The values of y that make these factors zero are -19 and 10. We know from the first term that (y-6) > 0, so y > 6. That means y = -19 is an extraneous solution.

The solution is ...

y = 10

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When solving equations using a graphing calculator, it often works well to define a function f(x) such that the solution is f(x) = 0, the x-intercept(s). That form is easily obtained by subtracting the right side of the equation from both sides of the equation. In part (a) here, that is ...

f(x) = e^(2x+1) -3e^(4-3x)