LOGARITHM MCQ LIST

 Question 1:

The value of $x$ satisfying ${\log }_{10}(2^x+x-41)=x(1-{\log }_{10}5)$is:

[1] 35

[2] 40

[3] 41

[4] 43

Answer & Solution

Option # 3

We have,

${\log }_{10}(2^x+x-41)=x(1-{\log }_{10}5)$

$\Rightarrow {\log }_{10}(2^x+x-41)=x{\log }_{10}2={\log }_{10}\left(2^x\right)$

$\Rightarrow 2^x+x-41=2^x\Rightarrow x=41.$

---

Question 2:
If the product of the roots of the equation, $x^{\left(\frac{3}{4}\right){\left({\log }_{2}x\right)}^2+{\log }_{2}x-\left(\frac{5}{4}\right)}=\sqrt{2}$ is $\frac{1}{\sqrt[b]{ba}}$ (where a, b $\in N$) then the value of $(a+b)$

[1] 0

[2] 1

[3] 18

[4] 19

Answer & Solution

Option # 4

Take log on both the sides with base 2

$(\frac{3}{4}{\left({\log }_{2}x\right)}^2+{\log }_{2}x-\frac{5}{4}){\log }_{2}x=\frac{1}{2}$

${\log }_{2}x=y$

$3y^3+4y^2-5y-2=0$

$\Rightarrow 3y^2(y-1)+7y(y-1)+2(y-1)=0$

$\Rightarrow (y-1)(3y^2+7y+2)=0$

$\Rightarrow (y-1)(3y+1)(y+2)=0$

$\Rightarrow y=1$ or $y=-2$ or $y=\frac{-1}{3}$

Therefore, $x=2;\frac{1}{4};\frac{1}{2^{1/3}}\Rightarrow x_1{x}_2{x}_3=\frac{1}{\sqrt[3]{316}}$

$\Rightarrow a+b=19$

---

Question 3:
For $0<a\ne 1,$ find the number of ordered pair $(x,y)$ satisfying the equation ${\log }_a|x+y|=\frac{1}{2}$ and ${\log }_{a}y-{\log }_a|x|={\log }_{a^2}4$

[1] 0

[2] 1

[3] 2

[4] 4

Answer & Solution

Option # 3

We have ${\log }_{a^2}|x+y|=\frac{1}{2}\Rightarrow |x+y|=a\Rightarrow x+y=\pm a\dots \left(1\right)$

Also, $\log \left(\frac{y}{|x|}\right)={\log }_{a^2}4\Rightarrow y=2|x|\dots \left(2\right)$

If $x>0,$ then $x=\frac{a}{3},y=\frac{2a}{3}$

If $x<0,$ then $y=2a,x=-a$

Therefore, possible ordered pairs $=(\frac{a}{3},\frac{2a}{3})$ and $(-a,2a)$

---

Question 4:
For $N>1,$ the product $\frac{1}{{\log }_{2}N}\cdot \frac{1}{{\log }_{N}8}\cdot \frac{1}{{\log }_{32}N}\cdot \frac{1}{{\log }_{N}128}$ simplifies to

[1] $\frac{3}{7}$

[2] $\frac{3}{7\ln 2}$

[3] $\frac{3}{5\ln 2}$

[4] $\frac{5}{21}$

Answer & Solution

Option # 4

$\frac{1}{{\log }_{2}N}\cdot \frac{1}{{\log }_{N}8}\cdot \frac{1}{{\log }_{32}N}\cdot \frac{1}{{\log }_{N}128}$

=$\frac{\ln 2}{\ln N}\cdot \frac{\ln N}{3\ln 2}\cdot \frac{5\ln 2}{\ln N}\cdot \frac{\ln N}{7\ln 2}$

 = $\frac{5}{21}$

---

Question 5:
If p is the smallest value of $x$ satisfying the equation $2^x+\frac{15}{2^x}=8$ then the value of $4^p$ is equal to

[1] 9

[2] 16

[3] 25

[4] 1

Answer & Solution

Option # 1

$2^{2x}-8\cdot 2^x+15=0$

$\Rightarrow (2^x-3)(2^x-5)=0$

$\Rightarrow 2^x=3$ or $2^x=5$

Hence smallest x is obtained by equating $2^x=3\Rightarrow x={\log }_{2}3$

So, $p={\log }_{2}3$

Hence, $4^p=2^{2{\log }_{2}3}=2^{{\log }_{2}9}=9$

---

Question 6:
The sum of two numbers a and b is $\sqrt{18}$ and their difference is $\sqrt{14}.$ The value of ${\log }_b$ a is equal to

[1] -1

[2] 2

[3] 1

[4] $\frac{1}{2}$

Answer & Solution

Option # 1

$a+b=\sqrt{18}$

$a-b=\sqrt{14}$

squaring  & subtract, we get $4ab=4\Rightarrow ab=1$

Hence number are reciprocal of each other $\Rightarrow {\log }_{b}a=-1$

---

Question 7:
The value of the expression ${\left({\log }_{10}2\right)}^3+{\log }_{10}8\cdot {\log }_{10}5+{\left({\log }_{10}5\right)}^3$ is

[1] rational which is less than 1

[2] rational which is greater than 1

[3] equal to 1

[4] an irrational number

Answer & Solution

Option # 3

${\log }_{10}2=a$ and ${\log }_{10}5=b$

$\Rightarrow a+b=1;a^3+3ab+b^3=?$

Now $(a+b{)}^3=1\Rightarrow a^3+b^3+3ab=1$

---

Question 8:
If $x=\frac{\sqrt{10}+\sqrt{2}}{2}$ and $y=\frac{\sqrt{10}-\sqrt{2}}{2},$ then the value of ${\log }_2(x^2+xy+y^2),$ is equal to

[1] 0

[2] 2

[3] 3

[4] 4

Answer & Solution

Option # 3

${\log }_2\left(\right(x+y{)}^2-xy)$

but $x+y=\sqrt{10};x-y=\sqrt{2};xy=\frac{10-2}{4}=2$

${\log }_2(10-2)={\log }_{2}8=3$

---

Question 9:
The value of $6+{\log }_{\frac{3}{2}}\left(\frac{1}{3\sqrt{2}}\sqrt{4-\frac{1}{3\sqrt{2}}\sqrt{4-\frac{1}{3\sqrt{2}}\sqrt{4-\frac{1}{3\sqrt{2}}\cdots }}}\right)$ is

[1] 1

[2] 2

[3] -4

[4] 4

Answer & Solution

Option # 4

Let $\sqrt{4-\frac{1}{3\sqrt{2}}\sqrt{4-\frac{1}{3\sqrt{2}}}}\dots \dots =t$

$\sqrt{4-\frac{1}{3\sqrt{2}}t}=t$

$4-\frac{1}{3\sqrt{2}}t=t^2$

$\Rightarrow t^2+\frac{1}{3\sqrt{2}}t-4=0$

$\Rightarrow 3\sqrt{2}{t}^2+t-12\sqrt{2}=0$

$t=\frac{-1\pm \sqrt{1+4\times 3\sqrt{2}\times 12\sqrt{2}}}{2\times 3\sqrt{2}}=\frac{-1\pm 17}{2\times 3\sqrt{2}}$

$t=\frac{16}{6\sqrt{2}},\frac{-18}{6\sqrt{2}}$

$t=\frac{8}{3\sqrt{2}},\frac{-3}{\sqrt{2}}$ and $\frac{-3}{\sqrt{2}}$ is rejected

So, $6+{\log }_{32}(\frac{1}{3\sqrt{2}}\times \frac{8}{3\sqrt{2}})$

$=6+{\log }_{32}\left(\frac{4}{9}\right)$

$=6+{\log }_{32}\left({\left(\frac{2}{3}\right)}^2\right)=6-2=4$

---

Question 10:
Suppose that $x<0.$ Which of the following is equal to $|2x-\sqrt{(x-2{)}^2}|$

[1] x-2

[2] 3x-2

[3] 3x+2

[4] -3x+2

Answer & Solution

Option # 4

$y=|2x-|x-2||=|2x-(2-x)|=|3x-2|$ as $x<0$ hence $y=2-3x$