Page 178 - Calculus Demystified

CHAPTER 6
SOLUTION Transcendental Functions 165
We take the natural logarithm of both sides:

4
x
3x
ln(5 · 2 ) = ln .
7 x
Applying the rules for logarithms we obtain
x
x
3x
ln(5 ) + ln(2 ) = ln 4 − ln(7 )
or
x · ln 5 + 3x · ln 2 = ln 4 − x · ln 7.
Gathering together all the terms involving x yields

x ·[ln 5 + 3 · ln 2 + ln 7]= ln 4
or
3
x ·[ln(5 · 2 · 7)]= ln 4.
Solving for x gives
ln 4
x = = log 280 4.
ln 280
EXAMPLE 6.23
Simplify the expression
5 · log 3 − (1/4) · log 16
7
7
B = .
3 · log 5 + (1/5) · log 32
7
7
SOLUTION
The numerator of B equals
5
log (3 ) − log (16 1/4 ) = log 243 − log 2 = log (243/2).
7
7
7
7
7
Similarly, the denominator can be rewritten as
3
log 5 + log (32 1/5 ) = log 125 + log 2 = log (125 · 2) = log 250.
7
7
7
7
7
7
Putting these two results together, we find that
log 243/2
7
B = = log 250 (243/2).
log 250
7
√
You Try It: What does 3 2 mean (in terms of the natural logarithm function)?
EXAMPLE 6.24
Simplify the expression (log 9) · (log 16).
4
9