ML Aggarwal Solution Class 9 Chapter 9 Logarithms Test

 Test

Question 1

Expand $\log _{a} \sqrt[3]{x^{7} y^{8} \div \sqrt[4]{z}}$

Sol :

⇒$\log _{a} \sqrt[3]{x^{7} y^{8} \div \sqrt[4]{z}}$

⇒$=\log _{a}\left(x^{7} y^{8} \div \sqrt[4]{z}\right)^{1 / 3}$

⇒$=\frac{1}{3} \log _{a}\left(x^{7} y^{8} \div \sqrt[4]{z}\right)$

⇒$=\frac{1}{3}\left[\log _{a} x^{7} y^{8}-\log _{a} \sqrt[4]{z}\right]$

⇒$=\frac{1}{3}\left[7 \log _{a} x+8 \log _{a} y-\log _{a}(z)^{1 / 4}\right]$

⇒$=\frac{1}{3}\left[7 \log _{a} x+8 \log _{a} y-\frac{1}{4} \log _{a} z\right]$

⇒$=\frac{7}{3} \log _{a} x+\frac{8}{3} \log _{a} y \div \frac{1}{12} \log _{a} z$


Question 2

Find the value of $\log \sqrt{3} 3 \sqrt{3}-\log _{5}(0.04)$

Sol :

⇒$\log _{\sqrt{3}} 3 \sqrt{3}-\log _{5}(0 \cdot 04)$

$=\log _{\sqrt{3}} 3+\log _{\sqrt{3}} \sqrt{3}-\log _{5} \frac{4}{100}$

$=\log _{\sqrt{3}} 3+1-\log _{5} \frac{1}{25}$

$=\log _{\sqrt{3}} 3+1-\log _{5} 5^{-2}$

$=\log _{\sqrt{3}} 3+1-(-2) \log _{5} 5$

$=\log _{\sqrt{3}}(\sqrt{3})^{2}+1+2 \times 1$

$=2 \log _{\sqrt{3}} \sqrt{3}+1+2$

=2×1+1+2

=2×1+2=5


Question 3

Prove the following:

(i) $(\log x)^{2}-(\log y)^{2}=\log \frac{x}{y} \cdot \log x y$

Sol :

⇒$(\log x)^{2}-(\log y)^{2}=\log \frac{x}{y} \cdot \log x y$

L.H.S$=(\log x)^{2}-(\log y)^{2}=(\log x-\log y)(\log x+\log y$

$\left[\because A^{2}-B^{2}=(A-B)(A+B)\right]$

$=\left(\log \frac{x}{y}\right)(\log x y)=\log \frac{x}{y} \cdot \log x y$=R.H.S

Result is proved


(ii) $2 \log \frac{11}{13}+\log \frac{130}{77}-\log \frac{55}{91}=\log 2 .$

Sol :

L.H.S$=2 \log \frac{11}{13}+\log \frac{130}{77}-\log \frac{55}{91}$

=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 

$=\log \left(\frac{10}{5}\right)=\log 2$

=R.H.S

Hence, Result is proved.


Question 4

If log (m + n) = log m + log n, show that $n=\frac{m}{m-1}$

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 $\log \frac{x+y}{2}=\frac{1}{2}(\log x+\log y),$ prove that x=y

Sol :

⇒$\log \frac{x+y}{2}=\frac{1}{2}(\log x+\log y)$

⇒$\log \frac{x+y}{2}=\frac{1}{2} \log (x \times y)$

⇒$\log \frac{x+y}{2}=\log (x \times y)^{\frac{1}{2}}$

Comparing, we get,

∴$\frac{x+y}{2}=(x \times y)^{\frac{1}{2}}=x y^{\frac{1}{2}} $

⇒$x+y=2(x y)^{\frac{1}{2}}$

Squaring

⇒$(x+y)^{2}=4 x y$

⇒$x^{2}+y^{2}+2 x y=4 x y$

⇒$x^{2}+y^{2}+2 x y-4 x y=0$

⇒$x^{2}+y^{2}-2 x y=0$

⇒$(x-y)^{2}=0$

⇒x-y=0

∴x=y 

Hence proved


Question 6

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

Sol :

⇒$a^{2}+b^{2}=27 a b$

⇒$a^{2}+b^{2}-2 a b=25 a b$

⇒$\frac{a^{2}+b^{2}-2 a b}{25}=a b$

⇒$\left(\frac{a-b}{5}\right)^{2}=a b$

Taking log both sides, $\log \left(\frac{a-b}{5}\right)^{2}=\log a b$

⇒$2 \log \left(\frac{a-b}{5}\right)=\log a+\log b$

⇒$\log \left(\frac{a-b}{5}\right)=\frac{1}{2}(\log a+\log b)$

Hence proved

Solve the following equations for x


Question 7

Solve the following equations for x:

(i) $\log _{x} \frac{1}{49}=-2$

⇒$(x)^{-2}=\frac{1}{49}$

⇒$(x)^{-2}=\left(\frac{1}{7}\right)^{2}$

⇒$(x)^{-2}=(7)^{-2}$

⇒x=7


(ii) $\log _{x} \frac{1}{4 \sqrt{2}}=-5$

Sol :

⇒$\frac{-1}{5} \log _{x} \frac{1}{4 \sqrt{2}}=1$

⇒$-\frac{1}{5} \log _{x} \frac{1}{\sqrt{32}}=1$

⇒$-\frac{1}{5} \log _{x} \frac{1}{2^{\frac{5}{2}}}=1$

⇒$-\frac{1}{5} \log _{x} 2^{\frac{-5}{2}}=1$

⇒$-\frac{1}{5} \times\left(\frac{-5}{2}\right) \log _{x} 2=1$

⇒$\frac{1}{2} \log _{x} 2=1$

⇒$\log _{x}\left(2^{\frac{1}{2}}\right)=1$

⇒$\log _{x} \sqrt{2}=\log _{x} x$

∴x=√2


(iii) $\log _{x} \frac{1}{243}=10$

⇒$\frac{1}{10} \log _{x} \frac{1}{243}=1$

⇒$\frac{1}{10} \log _{x} \frac{1}{3^{5}}=1$

⇒$\frac{1}{10} \log _{x}(3)^{-5}=1$

⇒$\frac{1}{10} \times(-5) \times \log _{x} 3=1$

⇒$-\frac{1}{2} \log _{x} 3=\log _{x} x$

⇒$\log _{x} 3^{-\frac{1}{2}}=\log _{x} x$

⇒$\log _{x} \frac{1}{\sqrt{3}}=\log _{x} x$

∴$x=\frac{1}{\sqrt{3}}$


(iv) $\log _{4} 32=x-4$

⇒$(4)^{x-4}=32$

⇒$\left[(2)^{2}\right]^{x-4}=2 \times 2 \times 2 \times 2 \times 2$

⇒$(2)^{2(x-4)}=(2)^{5}$

⇒$(2)^{2 x-8}=(2)^{5}$

⇒2x-8=5

⇒2x=5+8

⇒2x=13

⇒$x=\frac{13}{2}=6 \frac{1}{2}$


(v) $\log _{7}\left(2 x^{2}-1\right)=2$

⇒$(7)^{2}=2 x^{2}-1$

⇒$49=2 x^{2}-1$

⇒$50=2 x^{2}$

⇒$2 x^{2}=50$

⇒$x^{2}=\frac{50}{2}$

⇒$x^{2}=25$

⇒$x^{2}=\pm \sqrt{25}$

⇒x=+5,-5


(vi) $\log \left(x^{2}-21\right)=2$

⇒$(10)^{2}=x^{2}-21$

⇒$100=x^{2}-21$

⇒$x^{2}-21=100$

⇒$x^{2}=100+21$

⇒$x^{2}=121$

⇒$x=\pm \sqrt{121}$

⇒$x=\pm 11$

∴x=11,--11


(vii) $\log _{6}(x-2)(x+3)=1$

⇒$=1=\log _{6} 6$ $\left\{\because \log _{a} a=1\right\}$

Comparing :

⇒(x-2)(x+3)=6

⇒$x^{2}+3 x-2 x-6=6$

⇒$x^{2}+3 x-2 x-6=6$

⇒$x^{2}+x-6-6=0$

⇒$x^{2}+x-12=0$

⇒$x^{2}+4 x-3 x-12=0$

⇒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

Hence x=3, -4


(viii) $\log _{6}(x-2)+\log _{6}(x+3)=1$

⇒$\log _{6}(x-2)(x+3)=1=\log _{6} 6$ $\left\{\because \log _{a} a=1\right\}$

Comparing

⇒(x-2)(x+3)=6

⇒$x^{2}+3 x-2 x-6=6$

⇒$x^{2}+x-6-6=0$

⇒$x^{2}+x-12=0$

⇒$x^{2}+4 x-3 x-12=0$

⇒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

⇒$\log \left[(x+1)(x-1]=\log 11+\log (3)^{2}\right.$

⇒$\log \left(x^{2}-1\right)=\log 11+\log 9$ $\left[\because a^{2}-b^{2}=(a+b)(a-b)\right]$

⇒$\log \left(x^{2}-1\right)=\log (11 \times 9)$

⇒$x^{2}-1=11 \times 9$

⇒$x^{2}-1=99$

⇒$x^{2}=99+1$

⇒$x^{2}=100$

⇒$x^{2}=(10)^{2}$

⇒ x=16


Question 8

Solve for x and y:

$\frac{\log x}{3}=\frac{\log y}{2}$ and log (xy)=5

Sol :

⇒$\frac{\log x}{3}=\frac{\log y}{2}$

⇒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

⇒$\log x=\frac{15}{5}=3$

⇒$\frac{1}{3} \log x=1=\log 10$

⇒$\log x^{\frac{1}{3}}=\log 10$

∴$x^{\frac{1}{3}}=10$

⇒$x=10^{3}=1000$ (∵log 10=1)

Hence , x=1000

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

⇒3+log y=5

⇒log y=5-3=2

⇒$\frac{1}{2} \log y=1$

⇒$\log y^{\frac{1}{2}}=\log 10$ (∵log 10=1)

∴$y^{\frac{1}{2}}=10$

⇒$y=(10)^{2}=100$

Hence x=1000 and y=100


Question 9

If $a=1+\log _{x} y z$, $6=1+\log _{y} z x$ and $c=1+\log _{z} x y$, then show that ab+bc+ca=abc.

Sol :
⇒$a=1+\log _{x} y z$
⇒$b=1+\log _{y} z x$
⇒$c=1+\log _{z} x y$
⇒$a=1+\log _{x} y z=\log _{x} x+\log _{x} y z$

⇒$a=\log _{x} x y z$

⇒$\frac{1}{a}=\log _{x y z} x$

Similarly

$\frac{1}{b}=\log _{x y z} y$ and $\frac{1}{c}=\log _{x y z} z$


Now $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\log _{x g \varepsilon} x+\log _{x y z} y+\log _{x y z} z$

$=\frac{\log x}{\log _{x y z}}+\frac{\log y}{\log _{x y z}}+\frac{\log z}{\log _{x y z}}$

$=\frac{\log x+\log y+\log z}{\log _{x y z}}$

$=\frac{\log x y z}{\log _{x}}=1$

⇒$\frac{b c+c a+a b}{a b c}=1 $

⇒ab+bc+ca=abc

Hence proved

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