124 Basic Engineering Mathematics
Taking natural logs of both sides gives
ln
7
4
= ln e
3x
ln
7
4
= 3x ln e
Since ln e = 1, ln
7
4
= 3x
i.e.
0.55962 = 3x
i.e.
x = 0.1865,
correct to 4 significant figures.
Problem 16. Solve e x−1 = 2e 3x−4 correct to 4
significant figures
Taking natural logarithms of both sides gives
ln
e
x−1
= ln
2e
3x−4
and by the first law of logarithms,
ln
e
x−1
= ln 2 + ln
e
3x−4
i.e.
x − 1 = ln 2 + 3x − 4
Rearranging gives
4 − 1 − ln 2 = 3x − x
i.e.
3 − ln 2 = 2x
from which,
x =
3 − ln 2
2
= 1.153
Problem 17. Solve, correct to 4 significant
figures, ln(x − 2) 2 = ln(x − 2) − ln(x + 3) + 1.6
Rearranging gives
ln(x − 2)
2
− ln(x − 2) + ln(x + 3) = 1.6
and by the laws of logarithms,
ln
(x − 2) 2 (x + 3)
(x − 2)
= 1.6
Cancelling gives
ln{(x − 2)(x + 3)} = 1.6
and
(x − 2)(x + 3) = e
1.6
i.e.
x
2
+ x − 6 = e
1.6
or
x
2
+ x − 6 − e
1.6
= 0
i.e.
x
2
+ x − 10.953 = 0
Using the quadratic formula,
x =
−1 ±
1 2 − 4(1)(−10.953)
2
=
−1 ±
√
44.812
2
=
−1 ± 6.6942
2
i.e.
x = 2.847 or − 3.8471
x = −3.8471 is not valid since the logarithm of a
negative number has no real root.
Hence, the solution of the equation is x = 2.847
Now try the following Practice Exercise
Practice Exercise 65 Evaluating Napierian
logarithms (answers on page 347)
In problems 1 and 2, evaluate correct to 5 significant figures.
1. (a)
1
3
ln 5.2932
(b)
ln 82.473
4.829
(c)
5.62 ln 321.62
e 1.2942
2. (a)
1.786 lne 1.76
lg 10 1.41
(b)
5e −0.1629
2 ln0.00165
(c)
ln 4.8629 − ln 2.4711
5.173
In problems 3 to 16, solve the given equations, each
correct to 4 significant figures.
3. 1.5 = 4e 2t
4. 7.83 = 2.91e −1.7x
5. 16 = 24
1 − e
−
t
2
6. 5.17 = ln
x
4.64
7. 3.72 ln
1.59
x
= 2.43
8. ln x = 2.40
9. 24 + e 2x = 45
10. 5 = e x+1 − 7
Taking natural logs of both sides gives
ln
7
4
= ln e
3x
ln
7
4
= 3x ln e
Since ln e = 1, ln
7
4
= 3x
i.e.
0.55962 = 3x
i.e.
x = 0.1865,
correct to 4 significant figures.
Problem 16. Solve e x−1 = 2e 3x−4 correct to 4
significant figures
Taking natural logarithms of both sides gives
ln
e
x−1
= ln
2e
3x−4
and by the first law of logarithms,
ln
e
x−1
= ln 2 + ln
e
3x−4
i.e.
x − 1 = ln 2 + 3x − 4
Rearranging gives
4 − 1 − ln 2 = 3x − x
i.e.
3 − ln 2 = 2x
from which,
x =
3 − ln 2
2
= 1.153
Problem 17. Solve, correct to 4 significant
figures, ln(x − 2) 2 = ln(x − 2) − ln(x + 3) + 1.6
Rearranging gives
ln(x − 2)
2
− ln(x − 2) + ln(x + 3) = 1.6
and by the laws of logarithms,
ln
(x − 2) 2 (x + 3)
(x − 2)
= 1.6
Cancelling gives
ln{(x − 2)(x + 3)} = 1.6
and
(x − 2)(x + 3) = e
1.6
i.e.
x
2
+ x − 6 = e
1.6
or
x
2
+ x − 6 − e
1.6
= 0
i.e.
x
2
+ x − 10.953 = 0
Using the quadratic formula,
x =
−1 ±
1 2 − 4(1)(−10.953)
2
=
−1 ±
√
44.812
2
=
−1 ± 6.6942
2
i.e.
x = 2.847 or − 3.8471
x = −3.8471 is not valid since the logarithm of a
negative number has no real root.
Hence, the solution of the equation is x = 2.847
Now try the following Practice Exercise
Practice Exercise 65 Evaluating Napierian
logarithms (answers on page 347)
In problems 1 and 2, evaluate correct to 5 significant figures.
1. (a)
1
3
ln 5.2932
(b)
ln 82.473
4.829
(c)
5.62 ln 321.62
e 1.2942
2. (a)
1.786 lne 1.76
lg 10 1.41
(b)
5e −0.1629
2 ln0.00165
(c)
ln 4.8629 − ln 2.4711
5.173
In problems 3 to 16, solve the given equations, each
correct to 4 significant figures.
3. 1.5 = 4e 2t
4. 7.83 = 2.91e −1.7x
5. 16 = 24
1 − e
−
t
2
6. 5.17 = ln
x
4.64
7. 3.72 ln
1.59
x
= 2.43
8. ln x = 2.40
9. 24 + e 2x = 45
10. 5 = e x+1 − 7