ML Aggarwal Solution Class 9 Chapter 9 Logarithms Test
Test
Question 1
Expand loga3√x7y8÷4√z
Sol :
⇒loga3√x7y8÷4√z
⇒=loga(x7y8÷4√z)1/3
⇒=13loga(x7y8÷4√z)
⇒=13[logax7y8−loga4√z]
⇒=13[7logax+8logay−loga(z)1/4]
⇒=13[7logax+8logay−14logaz]
⇒=73logax+83logay÷112logaz
Question 2
Find the value of log√33√3−log5(0.04)
Sol :
⇒log√33√3−log5(0⋅04)
=log√33+log√3√3−log54100
=log√33+1−log5125
=log√33+1−log55−2
=log√33+1−(−2)log55
=log√3(√3)2+1+2×1
=2log√3√3+1+2
=2×1+1+2
=2×1+2=5
Question 3
Prove the following:
Sol :
⇒(logx)2−(logy)2=logxy⋅logxy
L.H.S=(logx)2−(logy)2=(logx−logy)(logx+logy
[∵A2−B2=(A−B)(A+B)]
=(logxy)(logxy)=logxy⋅logxy=R.H.S
Result is proved
(ii) 2log1113+log13077−log5591=log2.
Sol :
L.H.S=2log1113+log13077−log5591
=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=mm−1
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
∴x+y2=(x×y)12=xy12
⇒x+y=2(xy)12
Squaring
⇒(x+y)2=4xy
⇒x2+y2+2xy=4xy
⇒x2+y2+2xy−4xy=0
⇒x2+y2−2xy=0
⇒(x−y)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+b2−2ab=25ab
⇒a2+b2−2ab25=ab
⇒(a−b5)2=ab
Taking log both sides, log(a−b5)2=logab
⇒2log(a−b5)=loga+logb
⇒log(a−b5)=12(loga+logb)
Hence proved
Solve the following equations for x
Question 7
Solve the following equations for x:
⇒(x)−2=149
⇒(x)−2=(17)2
⇒(x)−2=(7)−2
⇒x=7
(ii) logx14√2=−5
Sol :
⇒−15logx14√2=1
⇒−15logx1√32=1
⇒−15logx1252=1
⇒−15logx2−52=1
⇒−15×(−52)logx2=1
⇒12logx2=1
⇒logx(212)=1
⇒logx√2=logxx
∴x=√2
(iii) logx1243=10
⇒110logx1243=1
⇒110logx135=1
⇒110logx(3)−5=1
⇒110×(−5)×logx3=1
⇒−12logx3=logxx
⇒logx3−12=logxx
⇒logx1√3=logxx
∴x=1√3
(iv) log432=x−4
⇒(4)x−4=32
⇒[(2)2]x−4=2×2×2×2×2
⇒(2)2(x−4)=(2)5
⇒(2)2x−8=(2)5
⇒2x-8=5
⇒2x=5+8
⇒2x=13
⇒x=132=612
(v) log7(2x2−1)=2
⇒(7)2=2x2−1
⇒49=2x2−1
⇒50=2x2
⇒2x2=50
⇒x2=502
⇒x2=25
⇒x2=±√25
⇒x=+5,-5
(vi) log(x2−21)=2
⇒(10)2=x2−21
⇒100=x2−21
⇒x2−21=100
⇒x2=100+21
⇒x2=121
⇒x=±√121
⇒x=±11
∴x=11,--11
(vii) log6(x−2)(x+3)=1
⇒=1=log66 {∵logaa=1}
Comparing :
⇒(x-2)(x+3)=6
⇒x2+3x−2x−6=6
⇒x2+3x−2x−6=6
⇒x2+x−6−6=0
⇒x2+x−12=0
⇒x2+4x−3x−12=0
⇒x(x+4)-3(x+4)=0
Either x+4=0, then x=-4
or x-3=0, then x=3
Hence x=3, -4
(viii) log6(x−2)+log6(x+3)=1
⇒log6(x−2)(x+3)=1=log66 {∵logaa=1}
Comparing
⇒(x-2)(x+3)=6
⇒x2+3x−2x−6=6
⇒x2+x−6−6=0
⇒x2+x−12=0
⇒x2+4x−3x−12=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)(x−1]=log11+log(3)2
⇒log(x2−1)=log11+log9 [∵a2−b2=(a+b)(a−b)]
⇒log(x2−1)=log(11×9)
⇒x2−1=11×9
⇒x2−1=99
⇒x2=99+1
⇒x2=100
⇒x2=(10)2
⇒ x=16
Question 8
Solve for x and y:
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.
⇒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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