Exponential functions 33
Taking natural logs of both sides gives:
ln
7
4
=lne^3 x
ln
7
4
= 3 xlne
Since lne=1ln
7
4
= 3 x
i.e. 0. 55962 = 3 x
i.e. x= 0. 1865 , correct to 4
significant figures.
Problem 15. Solve: ex−^1 =2e^3 x−^4 correct to 4
significant figures.
Taking natural logarithms of both sides gives:
ln
(
ex−^1
)
=ln
(
2e^3 x−^4
)
and by the first law of logarithms,
ln
(
ex−^1
)
=ln2+ln
(
e^3 x−^4
)
i.e. x− 1 =ln2+ 3 x− 4
Rearranging gives: 4− 1 −ln2= 3 x−x
i.e. 3 −ln2= 2 x
from which, x=
3 −ln2
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 ±
√
12 − 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 problemson
evaluating Napierian logarithms
In Problems 1 and 2, evaluate correct to 5 signifi-
cant figures:
- (a)
1
3
ln5.2932 (b)
ln82. 473
4. 829
(c)
5 .62ln321. 62
e^1.^2942
[(a) 0.55547 (b) 0.91374 (c) 8.8941]
- (a)
1 .786lne^1.^76
lg10^1.^41
(b)
5e−^0.^1629
2ln0. 00165
(c)
ln4. 8629 −ln2. 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.
- lnx= 2. 10 [8.166]
- 24+e^2 x= 45 [1.522]
- 5=ex+^1 −7[ 1 .485]
- 5 =4e^2 t [− 0 .4904]
- 83 = 2 .91e−^1.^7 x [− 0 .5822]
- 16= 24
(
1 −e−
t
2
)
[2.197]
- 17 =ln
( x
4. 64
)
[816.2]
- 3.72ln
(
1. 59
x
)
= 2. 43 [0.8274]