24 Higher Engineering Mathematics
1
2Hence, log4=logxlog√
becomes 4 =logx
i.e. log2=logx
from which, 2 =xi.e. thesolution of the equation is:x= 2Problem 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 logarithmslog(
x^2 − 3
x)
Hence, =log2x^2 − 3
xfrom which, = 2Rearranging gives: x^2 − 3 = 2 x
and x^2 − 2 x− 3 = 0
Factorizing gives: (x− 3 )(x+ 1 )= 0
from which, x=3orx=− 1x=−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= 3Now try the following exerciseExercise 12 Further problems on laws of
logarithmsIn 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
2log81+log36 [log 100]8.1
3log8−1
2log81+log27 [log 6]9.1
2log4−2log3+log45 [log 10]10.1
4log16+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
3log 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