24 Higher Engineering Mathematics
1
2
Hence, log4=logx
log
√
becomes 4 =logx
i.e. log2=logx
from which, 2 =x
i.e. thesolution of the equation is:x= 2
Problem 21. Solve the equation:
log
(
x^2 − 3
)
−logx=log2.
log
(
x^2 − 3
)
−logx=log
(
x^2 − 3
x
)
from the second law of logarithms
log
(
x^2 − 3
x
)
Hence, =log2
x^2 − 3
x
from which, = 2
Rearranging gives: x^2 − 3 = 2 x
and x^2 − 2 x− 3 = 0
Factorizing gives: (x− 3 )(x+ 1 )= 0
from which, x=3orx=− 1
x=−1 is not a valid solution since the logarithm of a
negative number has no real root.
Hence,the solution of the equation is:x= 3
Now try the following exercise
Exercise 12 Further problems on laws of
logarithms
In Problems 1 to 11, write as the logarithm of a
single number:
- log 2+log3 [log 6]
- log 3+log5 [log 15]
- log 3+log4−log6 [log 2]
- log 7+log21−log49 [log 3]
- 2log 2+log3 [log 12]
- 2log 2+3log5 [log 500]
- 2log 5−
1
2
log81+log36 [log 100]
8.
1
3
log8−
1
2
log81+log27 [log 6]
9.
1
2
log4−2log3+log45 [log 10]
10.
1
4
log16+2log3−log18 [log 1=0]
- 2log2+log5−log10 [log2]
Simplify the expressions given in Problems 12
to 14:
- log27−log9+log81
[log243 or log3^5 or 5log3] - log64+log32−log128
[log16 or log2^4 or 4log2] - log8−log4+log32
[log64 or log2^6 or 6log2]
Evaluate the expressions given in Problems 15
and 16:
15.
1
2 log16−
1
3 log8
log4
[0.5]
16.
log9−log3+^12 log81
2log3
[1.5]
Solve the equations given in Problems 17 to 22:
- logx^4 −logx^3 =log5x−log2x
[x= 2 .5] - log2t^3 −logt=log16+logt
[t=8] - 2logb^2 −3logb=log8b−log4b
[b=2] - log(x+ 1 )+log(x− 1 )=log3
[x=2]
1
3
log 27=log( 0. 5 a) [a=6]
- log
(
x^2 − 5
)
−logx=log4 [x=5]
3.3 Indicial equations
The laws of logarithms may be used to solve certain
equations involving powers—called indicial equa-
tions. For example, to solve, say, 3x=27, logari-
thms to a base of 10 are taken of both sides,
i.e. log 103 x=log 1027
and xlog 103 =log 10 27, by the third law of logarithms