Basic Engineering Mathematics, Fifth Edition

150 Basic Engineering Mathematics

aandb. Determine the cross-sectional area

needed for a resistance reading of 0.50ohms.

7. Corresponding experimental values of two
   quantitiesxandyare given below.

x 1.5 3.0 4.5 6.0 7.5 9.0

y 11.5 25.0 47.5 79.0 119.5 169.0

By plotting a suitable graph, verify thaty

andx are connected by a law of the form

y=kx^2 +c,wherek andc are constants.

Determine the law of the graph and hence find

the value ofxwhenyis 60.0

8. Experimental results of the safe loadLkN,
   applied to girders of varying spans,dm, are
   shown below.

Span,d(m) 2.0 2.8 3.6 4.2 4.8

Load,L(kN) 475 339 264 226 198

It isbelievedthat therelationshipbetweenload

and span isL=c/d,wherecis a constant.

Determine (a) the value of constantcand (b)

the safe load for a span of 3.0m.

9. The following results give corresponding val-
   ues of two quantities x and y which are
   believed to be related by a law of the form
   y=ax^2 +bx,whereaandbare constants.

x 33.8655.5472.8084.10111.4168.1

y 3.4 5.2 6.5 7.3 9.1 12.4

Verify the law and determine approximate val-

ues ofaandb. Hence, determine (a) the value

ofywhenxis 8.0 and (b) the value ofxwhen

yis 146.5

18.3 Revision of laws of logarithms

The laws of logarithms were stated in Chapter 15 as

follows:

log(A×B)=logA+logB (1)

log

(

A

B

)

=logA−logB (2)

logAn=n×logA (3)

Also, lne=1andif,say,lgx= 1. 5 ,

thenx= 101.^5 = 31. 62

Further, if 3x=7thenlg3x=lg 7 andxlg 3=lg 7,

from whichx=

lg 7

lg 3

= 1. 771

Theselawsandtechniquesareusedwhenevernon-linear

laws of the formy=axn,y=abxandy=aebxare

reduced to linear form withthevalues ofaandbneeding

to be calculated. This is demonstrated in the following

section.

18.4 Determination of laws involving

logarithms

Examples of the reduction of equations to linear form

involving logarithms include

(a) y=axn

Taking logarithms toabaseof10of both sides

gives

lgy=lg(axn)

=lga+lgxn by law (1)

i.e. lgy=nlgx+lgaby law (3)

which compares withY=mX+c

and shows thatlgyis plotted vertically against

lgxhorizontallyto produce a straight linegraph

of gradientnand lgy-axis intercept lg a.

See worked Problems 4 and 5 to demonstrate how

this law is determined.

(b) y=abx

Taking logarithms toabaseof10of both sides

gives

lgy=lg(abx)

i.e. lgy=lga+lgbx by law (1)

lgy=lga+xlgb by law (3)

i.e. lgy=xlgb+lga

or lgy=(lgb)x+lga

which compares with

Y=mX+c