Higher Engineering Mathematics, Sixth Edition

(Nancy Kaufman) #1

Logarithms 23


=6log2−7log2+5log2
by the third law of logarithms
=4log2

Problem 15. Write

1
2

log16+

1
3

log27−2log5
as the logarithm of a single number.

1
2

log16+

1
3

log27−2log5

=log16

1

(^2) +log27
1
(^3) −log5^2
by the third law of logarithms
=log

16 +log
√ 3
27 −log25
by the laws of indices
=log4+log3−log25
=log
(
4 × 3
25
)
by the first and second laws of logarithms
=log
(
12
25
)
=log0. 48
Problem 16. Write (a) log30 (b) log450 in terms
of log2,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+^12 log625
3log5
.
log25−log125+^12 log625
3log5


log5^2 −log5^3 +^12 log5^4
3log5


2log5−3log5+^42 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 third 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
log 4=logx.
1
2
log4=log4
1
(^2) from the third law of logarithms
=log

4 from the laws of indices

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