Download Some Solutions.......................................................................................... 1

Some Solutions..........................................................................................
1. This document is not complete. This is what is done NOW. MORE later.
2. The graph of y = ln(x – 2) + 3 is the same as that of y = lnx,
but shifted 2 units to the right, so its domain is (2,4 ) rather than (0,4 ),
then moved vertically 3 units (because y is 3 more than ln(x – 2) ).
x=2
x=0
The graph of y = ln|x| is the same as the graph of y = ln x where x is positive, and
where x is negative, it is just the reflection of y = ln x through the y-axis. y = ln |x –
1| is the graph of y = ln |x| , shifted
1 unit to the right, so the vertical asymptote occurs at x = 1, rather than at x = 0.
3.
Simplify an expression using properties of logarithms.
a.
Simplify completely: 2log5(5x) + log5(5x) – 3log5 x
2log5(5x) + log5(5x) – 3log5 x
log5(5x)2 + log5(5x) – log5 x3
(5x)2 (5x)
log5 ))))))))
x3
52 x2 5 x
log5 ))))))))
= log5 53 = 3
x3
b.
Simplify:
log2 (1/4) + log2 8 – log2 (3) + log2 (9) + log2 (.5)
–2
+
3
+ log2 )))
9
ln 3
+ –1 = log2 3 = ))))))
3
ln 2
4.
Solve a logarithmic equation.
log15 x + log15 (x – 2) = 1
log15 (x (x – 2)) = 1
x (x – 2)
So x = 5 or –3
= 15
x2 – 2x – 15 = 0
But a quick inspection shows
–3 is not a real solution,
because log functions are
not defined for negative input.
(x – 5) (x + 3) = 0
5.
Solve an exponential equation.
a.
Solve (answer must be exact): 2x+3 = 5x
2x+3 = 5x
Exponential; haul out the logs
ln( 2x+3) = ln (5x)
and use those log properties!
(x+3) ln 2
= x ln 5
x ln2 + 3ln2
= x ln 5
ln2 and ln5 are just numbers
3 ln2 = x (ln5 – ln2)
3 ln2
ln 5 – ln 2
)))))))))))
b.
Solve:
= x
2x = 3
ln 2x = ln3
x ln 2 = ln3
ln3
x = ))))) aka log
ln 2
6.
Solve a problem involving exponential growth.
a.
A bacterial population of 2400 has a relative growth rate of 4% per hour.
How long will it be until the population reaches 14,400?
Hmmm.... after 1 hour, there will be 2400(1+.04);
after two hours, the number should be 2400(1+.04)C(1+.04) again
After x hours, the number of bacteria will be 2400(1.04)x.
So this question asks at what value of x will:
2400(1.04)x = 14400
(1.04)x = 14400/2400 = 6
ln (1.04)x = ln 6
x ln (1.04) = ln 6
x = ln 6 / ln 1.04 = 45.68 (hours)
b.
If you deposit $5000 into an account earning interest at the annual rate of
5%, compounded annually, how long will it take to triple?
$5000 (1 + .05)t = $15000
(1 + .05)t =
ln(1.05)t
= ln 3
t ln(1.05)
= ln 3
t
c.
3
= ln 3/ ln1.05 Ñ 22.52 (years)
How does answer to (b) change if the interest is compounded... daily?
...Continuously?
After t years the value is $5000 (1 + .05/365)365t
Daily:
So, in a manner similar to above, we solve:
(1 + .05/365)365t = 3
And we get
365tCln (1 + .05/365)365t = ln 3
ln3
t =
))))))))))))))) Ñ
21.97 (years)
365Cln (1 + .05/365)
With continuous compounding, the computation is much neater:
The balance after t years at annual rate “r” for initial deposit P is
P e rt and we want that to be 3P.
We solve e rt = 3
Taking ln of both sides:
rt = ln 3 .... t = ln 4/r
In this case r = .05, so t = ln 3/.05 (years) Ñ 21.97 yrs