y(^0) x
(0,1)
(1,0)
y = ax
y^ = log
a^ x
y^ =
x
Figure 4.4The exponential and logarithm functions.
Note that as observed above, logaxdoes not exist for negative values ofx.
Solution to review question 4.1.6
A.We simply apply the laws of logarithms given above:
(i) lnx+2lny=lnx+lny^2
=ln(xy^2 )
(ii) 3 lnx−4lny=lnx^3 −lny^4
=ln(x^3 /y^4 )
(iii) 2 lnx−3ln( 2 x)+4lnx^3
=lnx^2 −ln( 8 x^3 )+lnx^12
=ln
(
x^2 ·x^12
8 x^3
)
=ln
(
x^11
8
)
(iv) 3 logax+2logax^2 =logax^3 +logax^4 =loga(x^3 ×x^4 )
=logax^7
(v) alogax+3loga(ax)=logaxa+loga(ax)^3
=logaxa+loga(a^3 x^3 )
=loga(xaa^3 x^3 )
=loga(a^3 x^3 +a)