A student had corrections to
do for a few questions on his logarithm test:
9) 9x+1 + 3x = 8
10) √ln
x + ln√x = 12
11) y = log4 x and x is
determined by rolling two dice and taking the sum. What is the probability that
y is rational?
12) log2x + log2y = 2, 2x·2y = 64,
find x2 + y2
Solutions:
9) 9x+1 + 3x
= 8
9x·91 + 3x = 8
9·9x + 3x = 8
9·(32)x + 3x = 8
9·32x + 3x = 8
9·(3x)2 + 3x = 8
Let u = 3x
9u2 + u = 8
9u2 + u – 8 = 0
(9u – 8)(u + 1) = 0
u = 8/9 or -1
3x = 8/9
x = log3 (8/9)
10) √ln
x + ln√x
= 12
I just started plugging in
values for x to get 12. 10 was too small, as was 100, 1,000, and so on up until
10,000,000. I found ln 10,000,000 to be just over 16, so I tried plugging in e16,
which worked.
√ln
x + ln√x
= 12
Let u = √ln
x
So u2 = ln x
u2/2 = 1/2·ln x = ln x1/2 = ln√x
which is the second term.
u + u2/2 = 12
The solutions to the
quadratic
u2/2 + u – 12 = 0
are u = 4 and -6.
4 = √ln
x
16 = ln x
e16
= x Answer
√ln
e16 + ln√
e16 = 12 it checks!
But u = -6 doesn’t:
-6 = √ln
x
36 = ln x
e36 = x
√ln e36 + ln√ e36 = 24
11) y = log4 x
For y to be rational, the only possible values for x
(that you
can get on two dice) are 2, 4 and 8. There's only 1 way to roll a two (1,1
"snake eyes"), 3 ways to roll a 4 (3,1)(1,3)(2,2), and 5 ways to roll
an 8 (2,6),(6,2),(3,5),(5,3) and (4,4). 9 ways out
of 36 possible outcomes, so the probability is 9/36 or 1/4.
12) log2x + log2y
= 2, 2x·2y = 64, find x2 + y2
log2(x·y) = 4 so
xy = 4
2x·2y
= 64 so 2x+y = 64 and x+y = 6
It never asks for x or y,
just x2 + y2
(x + y)2 = x2
+ y2 + 2xy
62 = x2 + y2 + 2(4)
36 = x2 + y2 + 8
28 = x2 + y2
The
unnecessarily involved algebraic solution
is
x = 4/y
4/y + y = 6
Multiply through by y
4 + y2 = 6y
Solve the quadratic
y2 – 6y + 4 = 0
y = 3 ± √5
When y = 3 + √5, x = 4/(3 + √5)
When y = 3 – √5, x = 4/(3 – √5)