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Solutions to a set of problems related to exponentials and logarithms. Topics covered include finding the value of logarithms, determining the domain and range of functions, and solving equations. The problems involve various calculations using logarithms and exponents.
Typology: Assignments
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2 2 2 2 2
2 2 2 2
2 ln( ) ln( 1) ln( 1) ln( ) ln( 1) ln( 1) ln[ ( 1)] ln( 1)
ln ln ln ( 1) 1 1
x x x x x x x x x
x x x x x x x x x
x − 3 > 0 ⇒ x > 3 So the damain is x ∈ (3, ∞)
And the range is y ∈ ℜ.
3.Find the exact value (no decimal answers) of
log (4) 3 log (5) 3 3
log (4) 3 log (5) 3 log (4 5) 3 log (20) 3
Solve for x in problems 4-
2 3 3 3
log ( 1) log ( 4) log 2 3 9 4 4
x x x x x x
x x x x x
Note that the domain of the original equation requires
x x x 1 x x
Obviously,
x = − is not in the domain, i.e., it is an extraneous solution. Hence the solution does not
exist for this equation.
1 3 6
1 2
ln 3 3 6 ( 1) ln 3 ln 6 (ln 6 ln 3) ln 3 ln(2) ln 3 log 3 ln 2
x x x x x x x
= ⇒ + = ⇒ − = ⇒ = ⇒ = =
. 4(3) 15 0
x − =
.09 .09. 3 3
4(3) 15 0 4(3) 15 (3) 0.09 log log 4 4 0.09 4
x x x x x
how long will it take for the investment to triple?
Compounded quarterly, so = 1
nt r A P n
, where r = 5.6% = 0.056, n = 4, A = 3 P , t = unkown
So
4 4 0.056 0.056 4 3 = 1+ 3 1+ 1.014 ln(3) 4 ln 1. 4 4
ln(3) 1.
4ln(1.014) 0.
t t t P P t
t
b. How long will it take if the interest is compounded continuously?
Compounded continuously, so , where 5.6% 0.056, 3
rt A = Pe r = = A = P
0.056 0.056 ln 3^ 1. 3 3 ln 3 0.056 19. 0.056 0.
t t ⇒ P = Pe = e ⇒ = t ⇒ t = ≈ ≈
this week. The number of people, N, who hear this rumor in t minutes is given by the formula,
0.15 t N N (^) f N ef
− = − , where f
N is the fixed population of the school. If the school has 1800 students and
staff members, how many minutes will it take for ¾ of the school to hear the rumor?
t t t t N (^) f N (^) f N ef N (^) f e e e
− − − − = − = − ⇒ = − ⇒ = − = =
ln(0.25) 1. 0.15 ln(0.25) 9. 0.15 0.
t t
half-life of Strontium-90 is 29 years. If 500 grams is present now, how much will be present in 50
years? (Hint: you need to use the formula , and you need to find first from the given
information)
kt p t = p e k
(29) 29 0 0
ln(0.5) 0. 0.5 0.5 29 ln(0.5) 0. 29 29
k k p p e e k k
0.0239(50) 1. p (50) 500 e 500 e 500(0.303) 151.
− − = = ≈ =
may be approximated by the formula log( E ) = 11.4 + (1.5) R. Find the energy that was released during
one of the biggest earthquakes in history which took place in 1933 in Japan and had a magnitude of 8.9.
log( E ) = 11.4 + (1.5) R = 11.4 + 1.5 × 8.9 = 11.4 + 13.35 = 24.75 ⇒ E = 10 ≈5.6E