114 Basic Engineering Mathematics

=log

(

4 × 3

25

)

by the first and second

laws of logarithms

=log

(

12

25

)

=log0. 48

Problem 16. Write (a) log 30 (b) log 450 in terms

oflog2,log3andlog5toanybase

(a) log30=log( 2 × 15 )=log( 2 × 3 × 5 )

=log2+log3+log5

by the first law

of logarithms

(b) log450=log( 2 × 225 )=log( 2 × 3 × 75 )

=log( 2 × 3 × 3 × 25 )

=log( 2 × 32 × 52 )

=log2+log3^2 +log5^2

by the first law

of logarithms

i.e. log450=log2+2log3+2log5

by the third law of logarithms

Problem 17. Write log

(

8 ×

√ 4

5

81

)

in terms of

log2,log3andlog5toanybase

log

(

8 ×^4

√

5

81

)

=log8+log^4

√

5 −log81

by the first and second laws

of logarithms

=log2^3 +log5

1

(^4) −log3^4
by the laws of indices
i.e. log
(
8 ×^4
√
5
81
)
=3log2+
1
4
log5−4log3
by the third law of logarithms
Problem 18. Evaluate
log25−log125+
1
2
log625
3log5
log25−log125+
1
2
log625
3log5

log5^2 −log5^3 +
1
2
log5^4
3log5

2log5−3log5+
4
2
log5
3log5

1log5
3log5

1
3
Problem 19. Solve the equation
log(x− 1 )+log(x+ 8 )=2log(x+ 2 )
LHS=log(x− 1 )+log(x+ 8 )=log(x− 1 )(x+ 8 )
from the first
law of logarithms
=log(x^2 + 7 x− 8 )
RHS=2log(x+ 2 )=log(x+ 2 )^2
from the first
law of logarithms
=log(x^2 + 4 x+ 4 )
Hence, log(x^2 + 7 x− 8 )=log(x^2 + 4 x+ 4 )
from which, x^2 + 7 x− 8 =x^2 + 4 x+ 4
i.e. 7 x− 8 = 4 x+ 4
i.e. 3 x= 12
and x= 4
Problem 20. Solve the equation
1
2
log4=logx
1
2
log4=log4
1
(^2) from the third law of
logarithms
=log
√
4 from the laws of indices
Hence,
1
2
log4=logx
becomes log
√
4 =logx
i.e. log2=logx
from which, 2 =x
i.e. the solution of the equation isx= 2.