ML Aggarwal Solution Class 9 Chapter 9 Logarithms MCQs

 MCQs

correct Solution from the given four options (1 to 7):

Question 1

If log√3 27 = x, then the value of x is

(a) 3

(b) 4

(c) 6

(d) 9

Sol :

⇒$\log _{\sqrt{3}} 27=x$

⇒$(\sqrt{3})^{x}=27$

⇒$(3)^{\frac{1}{2} x x}=3^{3}$

⇒$3^{\frac{x}{2}}=3^{3}$

⇒$\frac{x}{2}=3$

⇒x=6...(c)

Question 2

If log₅ (0.04) = x, then the value of x is

(a) 2

(b) 4

(c) -4

(d) -2

Sol :

⇒$\log _{5}(0.04)=x$

⇒$5^{x}=0.04=\frac{4}{100}=\frac{1}{25}=5^{-2}$

∴x=-2...(d)

Question 3

If log_0.5 64 = x, then the value of x is

(a) -4

(b) -6

(c) 4

(d) 6

Sol :

⇒$\log _{0.5} 64=x$

⇒$0.5^{x}=64$

⇒$=\left(\frac{1}{2}\right)^{x}=2^{6}$

⇒$2^{-x}=2^{6}$

∴-x=6⇒x=-6...(b)

Question 4

If $\log _{10} \sqrt[3]{5} x=-3$, then the value of x is

(a) $\frac{1}{5}$

(b) $-\frac{1}{5}$

(c) -1

(d) 5

Sol :

$\log _{\sqrt[3]{5}} x=-3,(\sqrt[3]{5})^{-3}=x$

$x=\left(5^{\frac{1}{3}}\right)^{-3}=5^{\frac{1}{3}(-3)}=5^{-1}$

$x=\frac{1}{5}$

(b)

Question 5

If log (3x + 1) = 2, then the value of x is

(a) $\frac{1}{3}$

(b) 99

(c) 33

(d) $\frac{19}{3}$

Sol :

⇒log (3x+1)=2=log 100 (∵ log 100=2)

∴3x+1=100

⇒3x=100-1=99

⇒$x=\frac{99}{3}=33$

(c)

Question 6

The value of $2+\log _{10}(0.01)$ is

(a) 4

(b) 3

(c) 1

(d) 0

Sol :

⇒$2+\log _{10}(0.01)$

⇒2+(-20)=2-2=0

(d)

Question 7

The value of $\frac{\log 8-\log 2}{\log 32}$ is

(a) $\frac{2}{5}$

(b) $\frac{1}{4}$

(c) $-\frac{2}{5}$

(d) $\frac{1}{3}$

Sol :

⇒$\frac{\log 8-\log 2}{\log 32}=\frac{\log \frac{8}{2}}{\log 2^{5}}$

⇒$=\frac{\log 4}{\log 2^{5}}=\frac{\log 2^{2}}{\log 2^{5}}$

⇒$=\frac{2 \log 2}{5 \log 2}=\frac{2}{5}$

(a)