One week a student exercised 3 hours at school and another of an hour at home. If of the student's total exercise came from play
1 answer:
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Answer:
(a) x = (3 -ln(3))/5 ≈ 0.819722457734
(b) y = 10
Step-by-step 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)
