Understanding Engineering Mathematics

B.If log 2 x=6 then from the change of base formula

logax=

logbx

logba

we have

log 8 x=

log 2 x

log 28

=

log 2 x

3

=

6

3

= 2

Alternatively, if log 2 x=6thenx= 26 =( 23 )^2 = 82 ,solog 8 x=2,

as above.

C.lny=2lnx−^1 +ln(x− 1 )+ln(x+ 1 )

=lnx−^2 +ln(x− 1 )+ln(x+ 1 )

=ln(x−^2 (x− 1 )(x+ 1 ))

=ln

(

x^2 − 1

x^2

)

So y=

x^2 − 1

x^2

4.2.7 Some applications of logarithms

➤

120 138➤

When an unknown occurs in an index in an equation, as for example in

ax=b

then we may be able to solve the equation by taking logs. If the baseahappened to be
ethen, of course, the ‘natural’ thing to do would be to take natural logs, but we can use
the same idea whatever the base. Thus, taking natural logs we obtain

ln(ax)=xlna=lnb
so

x=

lnb

lna

Conversely, if an unknown occurs in a logarithm, then we can sometimes solve by taking
an exponential. For example, the equation

log 2 (x+ 1 )= 3 +log 2 x

can be solved by first gathering the logs together to give

log 2 (x+ 1 )−log 2 x=log 2

[

x+ 1

x

]

= 3

We can now remove the log by exponentiating with 2 – i.e. raising 2 to the power of each
side to give (2 to the power of log 2 xisx)

x+ 1

x

= 23 = 8

from which we findx=^17.