Exponential functions 33
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 15. 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 16. 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 exercise
Exercise 17 Further problems on
evaluating Napierian logarithms
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
[(a) 0.55547 (b) 0.91374 (c) 8.8941]
2. (a)
1.786 ln e 1.76
lg 10 1.41
(b)
5e −0.1629
2 ln0.00165
(c)
ln 4.8629 − ln 2.4711
5.173
[(a) 2.2293 (b) −0.33154 (c) 0.13087]
In Problems 3 to 7 solve the given equations, each
correct to 4 significant figures.
3. ln x = 2.10
[8.166]
4. 24 + e
2x
= 45
[1.522]
5. 5 = e x+1 − 7
[ 1 .485]
6. 1.5 = 4e 2t
[−0.4904]
7. 7.83 = 2.91e −1.7x
[−0.5822]
8. 16 = 24
1 − e
−
t
2
[2.197]
9. 5.17 = ln
x
4.64
[816.2]
10. 3.72 ln
1.59
x
= 2.43
[0.8274]
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 15. 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 16. 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 exercise
Exercise 17 Further problems on
evaluating Napierian logarithms
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
[(a) 0.55547 (b) 0.91374 (c) 8.8941]
2. (a)
1.786 ln e 1.76
lg 10 1.41
(b)
5e −0.1629
2 ln0.00165
(c)
ln 4.8629 − ln 2.4711
5.173
[(a) 2.2293 (b) −0.33154 (c) 0.13087]
In Problems 3 to 7 solve the given equations, each
correct to 4 significant figures.
3. ln x = 2.10
[8.166]
4. 24 + e
2x
= 45
[1.522]
5. 5 = e x+1 − 7
[ 1 .485]
6. 1.5 = 4e 2t
[−0.4904]
7. 7.83 = 2.91e −1.7x
[−0.5822]
8. 16 = 24
1 − e
−
t
2
[2.197]
9. 5.17 = ln
x
4.64
[816.2]
10. 3.72 ln
1.59
x
= 2.43
[0.8274]