ML Aggarwal Solution Class 9 Chapter 9 Logarithms Test

 Test

Question 1

Expand loga3x7y8÷4z

Sol :

loga3x7y8÷4z

=loga(x7y8÷4z)1/3

=13loga(x7y8÷4z)

=13[logax7y8loga4z]

=13[7logax+8logayloga(z)1/4]

=13[7logax+8logay14logaz]

=73logax+83logay÷112logaz


Question 2

Find the value of log333log5(0.04)

Sol :

log333log5(004)

=log33+log33log54100

=log33+1log5125

=log33+1log552

=log33+1(2)log55

=log3(3)2+1+2×1

=2log33+1+2

=2×1+1+2

=2×1+2=5


Question 3

Prove the following:

(i) (logx)2(logy)2=logxylogxy

Sol :

(logx)2(logy)2=logxylogxy

L.H.S=(logx)2(logy)2=(logxlogy)(logx+logy

[A2B2=(AB)(A+B)]

=(logxy)(logxy)=logxylogxy=R.H.S

Result is proved


(ii) 2log1113+log13077log5591=log2.

Sol :

L.H.S=2log1113+log13077log5591

=2[log 11-log 13]+[log 130- log 77]-[log 55- log 91]

=2[log 11-log 13]+[log 13×10-log 11×7]-[log 11×5-log 13×7]

=2[log 11-log 13]+[(log 13+log 10)-(log 11+log 7)]-[(log 11+log 5)-(log 13+log 7)]

=2 log 11-2 log 13+ log 10- log 11- log 7-log 11- log 5+ log 13+ log 7

=(2 log 11-log 11-log 11)+(-2 log 13+log 13+log 13)+log 10-log 5+(log 7-log 7)

=0+0+ log 10- log 5+0= log 10- log 5 

=log(105)=log2

=R.H.S

Hence, Result is proved.


Question 4

If log (m + n) = log m + log n, show that n=mm1

Sol :

Given log (m+n)= log m+log n

⇒log (m+n)=\log m n 

⇒m+n=m n

⇒m=m n-n 

⇒m=n(m-1)

⇒n(m-1)=m 

⇒$n=\frac{m}{m-1}

Hence, result is proved.


Question 5

If logx+y2=12(logx+logy), prove that x=y

Sol :

logx+y2=12(logx+logy)

logx+y2=12log(x×y)

logx+y2=log(x×y)12

Comparing, we get,

x+y2=(x×y)12=xy12

x+y=2(xy)12

Squaring

(x+y)2=4xy

x2+y2+2xy=4xy

x2+y2+2xy4xy=0

x2+y22xy=0

(xy)2=0

⇒x-y=0

∴x=y 

Hence proved


Question 6

If  a, b are positive real numbers, a>b and a2+b2=27ab, prove that \log \left(\frac{a-b}{5}\right)=\frac{1}{2}(\log a+\log b)

Sol :

a2+b2=27ab

a2+b22ab=25ab

a2+b22ab25=ab

(ab5)2=ab

Taking log both sides, log(ab5)2=logab

2log(ab5)=loga+logb

log(ab5)=12(loga+logb)

Hence proved

Solve the following equations for x


Question 7

Solve the following equations for x:

(i) logx149=2

(x)2=149

(x)2=(17)2

(x)2=(7)2

⇒x=7


(ii) logx142=5

Sol :

15logx142=1

15logx132=1

15logx1252=1

15logx252=1

15×(52)logx2=1

12logx2=1

logx(212)=1

logx2=logxx

∴x=√2


(iii) logx1243=10

110logx1243=1

110logx135=1

110logx(3)5=1

110×(5)×logx3=1

12logx3=logxx

logx312=logxx

logx13=logxx

x=13


(iv) log432=x4

(4)x4=32

[(2)2]x4=2×2×2×2×2

(2)2(x4)=(2)5

(2)2x8=(2)5

⇒2x-8=5

⇒2x=5+8

⇒2x=13

x=132=612


(v) log7(2x21)=2

(7)2=2x21

49=2x21

50=2x2

2x2=50

x2=502

x2=25

x2=±25

⇒x=+5,-5


(vi) log(x221)=2

(10)2=x221

100=x221

x221=100

x2=100+21

x2=121

x=±121

x=±11

∴x=11,--11


(vii) log6(x2)(x+3)=1

=1=log66 {logaa=1}

Comparing :

⇒(x-2)(x+3)=6

x2+3x2x6=6

x2+3x2x6=6

x2+x66=0

x2+x12=0

x2+4x3x12=0

⇒x(x+4)-3(x+4)=0

⇒(x+4)(x-3)=0

Either x+4=0, then x=-4

or x-3=0, then x=3

Hence x=3, -4


(viii) log6(x2)+log6(x+3)=1

log6(x2)(x+3)=1=log66 {logaa=1}

Comparing

⇒(x-2)(x+3)=6

x2+3x2x6=6

x2+x66=0

x2+x12=0

x2+4x3x12=0

⇒x(x+4)-3(x+4)=0

⇒(x+4)(x-3)=0

Either x+4=0, then x=-4

or x-3=0, then x=3

∴x=3 ,4


(ix) log(x+1)+log(x-1)=log 11+2 log 3

log[(x+1)(x1]=log11+log(3)2

log(x21)=log11+log9 [a2b2=(a+b)(ab)]

log(x21)=log(11×9)

x21=11×9

x21=99

x2=99+1

x2=100

x2=(10)2

⇒ x=16


Question 8

Solve for x and y:

logx3=logy2 and log (xy)=5

Sol :

logx3=logy2

⇒2 log x=3 log y

⇒2 log x-3 log y=0..(i)

and  log xy=5

⇒log x+log y=5..(ii)

Multiplying (ii) by 3 and (i) by 1

 ⇒2 log x-3 log y=0

⇒3 log x+3 log y=15

Adding  5 log x=15

logx=155=3

13logx=1=log10

logx13=log10

x13=10

x=103=1000 (∵log 10=1)

Hence , x=1000

Substituting the value of log x=3 in (ii)

⇒3+log y=5

⇒log y=5-3=2

12logy=1

logy12=log10 (∵log 10=1)

y12=10

y=(10)2=100

Hence x=1000 and y=100


Question 9

If a=1+logxyz, 6=1+logyzx and c=1+logzxy, then show that ab+bc+ca=abc.

Sol :
a=1+logxyz
b=1+logyzx
c=1+logzxy
a=1+logxyz=logxx+logxyz

a=logxxyz

1a=logxyzx

Similarly

1b=logxyzy and 1c=logxyzz


Now 1a+1b+1c=logxgεx+logxyzy+logxyzz

=logxlogxyz+logylogxyz+logzlogxyz

=logx+logy+logzlogxyz

=logxyzlogx=1

bc+ca+ababc=1

⇒ab+bc+ca=abc

Hence proved

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