Logarithm and Antilogarithm Test 19

Result

Logarithm and Antilogarithm Test 19

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

$$\log {3} + \log {5}$$ is equal to..........

A
$$\log {15}$$

B
$$\log {8}$$

C
$$\dfrac{\log {3}}{\log{5}}$$

D
$$\log{2}$$

Solution

$$\textbf{Step 1 : Use the properties of log }$$

$$\text{Given , }\log 3+\log5$$

$$=\log(3\times5)$$ [ $$\boldsymbol{\because \log a+\log b =\log(a \times b)}$$]

$$=\log 15$$

$$\textbf{Hence, A is the correct option.}$$
Question 2
1 / -0

Simplify the following using law of exponents for $$a=2, x=1,y=1,z=1$$
$$a^x\times a^y\times a^z$$

A
$$2$$

B
$$4$$

C
$$8$$

D
$$16$$

Solution

we know,

$$a^{x}*a^{y}*a^{z}=a^{x+y+z}$$

$$\therefore2^{1+1+1}$$

$$=2^{3}$$

$$=8$$
Question 3
1 / -0

If $$x^2+y^2=25$$ , then $$log_5 \begin {bmatrix} Max (3x+4y) \end {bmatrix}$$ is

A
$$2$$

B
$$3$$

C
$$4$$

D
$$5$$

Solution

$$log_5(3x+4y)$$

Let $$s=3x+4y$$
given $$x^2+y^2=25$$
then $$S=3x+4 \sqrt{25-x^2}$$
$$\dfrac{ds}{dx}=3+4 \dfrac{1}{2\sqrt(25-x^2)}$$
$$3=\dfrac{4x}{\sqrt{25-x^2}}$$
$$9(25-x^2)=16x^2$$
$$x=\pm 3$$

and $$x^2+y^2=25$$
$$y^2=25-x^2$$
$$y=\pm 4$$
$$\dfrac{d^2s}{dx^2}<0$$ ; At $$x=3 \,and\, y=4$$

$$S=3x+4y=3(3)+4(4)=25$$
$$log_5 (3x+4y)=log_5(s)=log_5(25)=log_5(5^2)=2$$
Question 4
1 / -0

Simplify the following using law of exponents.
$$(-3)^3\times (-3)^{10}\times (-3)^7$$

A
$$3^{20}$$

B
$$3^{40}$$

C
$$3^{60}$$

D
$$3^{10}$$

Solution

we know,

$$a^{m}*a^{n}*a^{p}=a^{m+n+p}$$

so,

$$(-3)^{3}*(-3)^{10}*(-3)^{7}=(-3)^{3+10+7}$$

$$=(-3)^{20}$$

$$=(3)^{20}$$
Question 5
1 / -0

$$\log_8 {64}$$ is equal to_____

A
$$2$$

B
$$3$$

C
$$4$$

D
$$5$$

Solution

$$\log_8(64)$$
$$=\log_8(8^2)$$
$$=2\log_88$$
$$=2$$ ($$\because \log_aa=1)$$
Hence, A is the correct option.
Question 6
1 / -0

$$\log_4{64}$$ is equal to____

A
$$2$$

B
$$3$$

C
$$4$$

D
$$5$$

Solution

$$\log_4(64)$$
$$=\log_4(4^3)$$
$$=3\log_44$$
$$=3$$ ($$\because \log_aa=1)$$
Hence, B is the correct option.
Question 7
1 / -0

If $$\displaystyle \int{ \frac{(\sqrt{x})^5}{(\sqrt{x})^7 + x^6} } dx = a\ log \left( \frac{x^k}{1 + x^k} \right) + c$$, then a and k are

A
$$\displaystyle \frac{2}{5}, \frac{5}{2}$$

B
$$\displaystyle \frac{1}{5}, \frac{2}{5}$$

C
$$\displaystyle \frac{5}{2}, \frac{1}{2}$$

D
$$\displaystyle \frac{2}{5}, \frac{1}{2}$$

Solution

Given
$$ \int \dfrac{(\sqrt{x})^5}{(\sqrt{x})^7+x^6} dx = a log \left ( \dfrac{x^k}{1+x^k} \right )+c $$
L.H.S Taking $$ x^6 $$ common from denominator
$$ \displaystyle \Rightarrow \int \dfrac{(\sqrt{x})^5}{x^6 \left ( \dfrac{(\sqrt{x})^7}{x^6}+1 \right )}dx$$
$$ \displaystyle \Rightarrow \int \dfrac{1}{(\sqrt{x})^7 (\dfrac{1}{\sqrt{x}})^5 + 1 }dx$$
Now Let $$ \dfrac{1}{(\sqrt{x})^5} = t $$
$$ \dfrac{-5}{2(\sqrt{x})^7} .dx = dt $$
$$\displaystyle \Rightarrow \int \frac{-2}{3}.\frac{1}{(t+1)}dt $$
$$ \Rightarrow \dfrac{-2}{5} ln (t+1)+c $$
put $$ t = \dfrac{1}{(\sqrt{x})^5} $$
$$ \Rightarrow \dfrac{-2}{5} ln \left ( \dfrac{1+(\sqrt{x})^5}{(\sqrt{x})^5} \right )+c \Rightarrow \dfrac{2}{5} ln \left ( \dfrac{(\sqrt{x})^5}{1+(\sqrt{x})^5} \right ) +c $$
on comparing $$ \boxed{a = t\dfrac{2}{5}} \boxed {k = 5/2}$$
Question 8
1 / -0

If the mantissa of $$\log 2125 =3.3273$$, find the mantissa of $$\log21.25$$

A
$$1.3273$$

B
$$2.3273$$

C
$$0.3273$$

D
$$32.2321$$

Solution
Question 9
1 / -0

$$\log_3 {27}$$ is equal to____

A
$$3$$

B
$$2$$

C
$$4$$

D
$$5$$

Solution

$$\textbf{Step 1 : Using the principle properties of logarithm}$$
$$\boldsymbol{\log_a m^n = n \log_a m\, ; \, where \,\,m, a > 0,\neq1}$$

$$\text{Given : }\log_3 27$$

$$= \log_3 3^3$$

$$\textbf{Step 2 : Using the principle properties of logarithm}$$
$$\boldsymbol{\log_a a = 1 ; \, where \,\,a > 0,\neq1}$$

$$= 3 \log_3 3 = 3$$
Question 10
1 / -0

$$\log {5} + \log {8}$$ is equal to........

A
$$\log {40}$$

B
$$\log{13}$$

C
$$\dfrac{\log{8}}{\log{5}}$$

D
$$\log{3}$$

Solution

$$\textbf{Step 1 : Use properties of logarithm }$$

$$\text{Given , }\log 5+\log8$$

$$=\log(5\times8)$$ $$\boldsymbol{[\because \log a+\log b =\log(a \times b)]}$$

$$=\log 40$$

$$\textbf{Hence, A is the correct option.}$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Logarithm and Antilogarithm Test 19

Result

Logarithm and Antilogarithm Test 19

Score

-

Rank

-

SHARING IS CARING

Question 1

$$\log {15}$$

$$\log {8}$$

$$\dfrac{\log {3}}{\log{5}}$$

$$\log{2}$$

Solution

Question 2

$$2$$

$$4$$

$$8$$

$$16$$

Solution

Question 3

$$2$$

$$3$$

$$4$$

$$5$$

Solution

Question 4

$$3^{20}$$

$$3^{40}$$

$$3^{60}$$

$$3^{10}$$

Solution

Question 5

$$2$$

$$3$$

$$4$$

$$5$$

Solution

Question 6

$$2$$

$$3$$

$$4$$

$$5$$

Solution

Question 7

$$\displaystyle \frac{2}{5}, \frac{5}{2}$$

$$\displaystyle \frac{1}{5}, \frac{2}{5}$$

$$\displaystyle \frac{5}{2}, \frac{1}{2}$$

$$\displaystyle \frac{2}{5}, \frac{1}{2}$$

Solution

Question 8

$$1.3273$$

$$2.3273$$

$$0.3273$$

$$32.2321$$

Solution

Question 9

$$3$$

$$2$$

$$4$$

$$5$$

Solution

Question 10

$$\log {40}$$

$$\log{13}$$

$$\dfrac{\log{8}}{\log{5}}$$

$$\log{3}$$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.