Logarithmic Function MCQ Quiz - Objective Question with Answer for Logarithmic Function - Download Free PDF

Concept:

• We are given:
  • t = e^2x
  • y = log_e(t²) = 2 log_e(t)
• We need to find the second derivative d²y/dx².
• This requires the chain rule and product rule of differentiation.

Calculation:

Step 1: Express y in terms of x

y = 2 log(t), where t = e^2x

Since log(t) = log(e^2x) = 2x

⇒ y = 2 × 2x = 4x

First derivative:

dy/dx = d/dx (4x) = 4

Second derivative:

d²y/dx² = d/dx (4) = 0

∴ The correct answer is: 0.

Let, $\mathrm{x}={\mathrm{y}}^2={\mathrm{z}}^3={\mathrm{w}}^4={\mathrm{u}}^5=\mathrm{k}$

$\mathrm{x}=\mathrm{k},\mathrm{y}={\mathrm{k}}^{1/2},\mathrm{z}={\mathrm{k}}^{1/3},\mathrm{w}={\mathrm{k}}^{1/4},\mathrm{u}={\mathrm{k}}^{1/5}$

Therefore, $\mathrm{x}\mathrm{y}\mathrm{z}\mathrm{w}\mathrm{u}={\mathrm{k}}^{1+1/2+1/3+1/4+1/5}={\mathrm{k}}^{137,60}$

Therefore, ${\log }_{\mathrm{x}}{}^{\mathrm{x}\mathrm{y}\mathrm{z}\mathrm{w}\mathrm{u}}={\log }_{\mathrm{k}}{\mathrm{k}}^{13750}=2\frac{17}{60}$

Calculation

Given equation is ${\log }_{10}\left({\displaystyle \frac{x^3-y^3}{x^3+y^3}}\right)=2$

${\displaystyle \frac{x^3-y^3}{x^3+y^3}}=100$

applying componendo and dividendo gives,

$\Rightarrow -{\displaystyle \frac{x^3}{y^3}}={\displaystyle \frac{101}{99}}$

$\Rightarrow {\displaystyle \frac{x^3}{y^3}}=-{\displaystyle \frac{101}{99}}$

Differentiating with respect to $x$ gives,

$⟹3{\displaystyle \frac{x^2}{y^2}}\left({\displaystyle \frac{y-x{\displaystyle \frac{dy}{dx}}}{y^2}}\right)=0$

$\therefore {\displaystyle \frac{dy}{dx}}={\displaystyle \frac{y}{x}}$

Hence option 4 is correct

Calculation

y = logn x

$\frac{dy}{dx}=\frac{1}{(xlo{g}^{n-1}xlo{g}^{n-2}x...logx)}$

x log x log2 x log3 x ..... logn-1 x logn x $\frac{dy}{dx}$ = logn x

Hence option 4 is correct 

Answer : 1

Solution :

log(x + y) = 2xy

Differentiating w.r.t. x, we get

$\frac{1}{x+y}(1+\frac{\mathrm{d}y}{\mathrm{d}x})=2x\frac{\mathrm{d}y}{\mathrm{d}x}+2y$

$\frac{1}{x+y}(1+\frac{\mathrm{d}y}{\mathrm{d}x})=2x\frac{\mathrm{d}y}{\mathrm{d}x}+2y$

$(\frac{1}{x+y}-2x)\frac{\mathrm{d}y}{\mathrm{d}x}=2y-\frac{1}{x+y}$

$\frac{\mathrm{d}y}{\mathrm{d}x}(\frac{1}{x+y}-2x)=2y-\frac{1}{x+y}$

$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{(2y-\frac{1}{x+y})}{(\frac{1}{x+y}-2x)}$

For x = 0, log(y) = 0

⇒ y = 1

${\frac{\mathrm{d}y}{\mathrm{d}x}|}_{(0,1)}=\frac{(2-\frac{1}{0+1})}{(\frac{1}{0+1}-0)}=1$

Concept:

Formula:

log mn = n log m

$\frac{\mathrm{d}(\mathrm{u}\mathrm{v})}{\mathrm{d}\mathrm{x}}=\mathrm{v}\frac{\mathrm{d}\mathrm{u}}{\mathrm{d}\mathrm{x}}+\mathrm{u}\frac{\mathrm{d}\mathrm{v}}{\mathrm{d}\mathrm{x}}$

$\frac{\mathrm{d}\log \mathrm{x}}{\mathrm{d}\mathrm{x}}=\frac{1}{\mathrm{x}}$

$\frac{\mathrm{d}\mathrm{x}}{\mathrm{d}\mathrm{x}}=1$

Calculation:

Let y = x^log x

Taking log both sides, we get

⇒ log y = x^{log x}

⇒ log y = log x log  x            (∵ log mn = n log m)

Differentiating with respect to x, we get

⇒ $\frac{1}{\mathrm{y}}\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=\log \mathrm{x}\frac{\mathrm{d}\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{x}}{\mathrm{d}\mathrm{x}}+\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{x}\frac{\mathrm{d}\log \mathrm{x}}{\mathrm{d}\mathrm{x}}$

⇒ $\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=\mathrm{y}(\log \mathrm{x}\times \frac{1}{\mathrm{x}}+\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{x}\times \frac{1}{\mathrm{x}})$

⇒ $\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=\frac{{\mathrm{x}}^{\log \mathrm{x}}}{\mathrm{x}}(2\log \mathrm{x})$

⇒ $\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}={\mathrm{x}}^{\log \mathrm{x}-1}(\log {\mathrm{x}}^2)$

Concept:

Product rule of differentiation:

Let h(x) = f(x)g(x), then h'(x) = f'(x)g(x) + f(x)g'(x)

Calculation:

Given, y = logy x

⇒ y = $\frac{\log x}{\log y}$

⇒ y log y = log x

By differentiating both sides w.r.t. x, we get

$\frac{dy}{dx}\cdot \log y+y\cdot \frac{1}{y}\cdot \frac{dy}{dx}=\frac{1}{x}$

$\log y\frac{dy}{dx}+\frac{dy}{dx}=\frac{1}{x}$

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

∴ $\frac{dy}{dx}=\frac{1}{x(1+\log y)}$

Given:

log 2 =  x, log 3 = y

Concept:

Properties of Logarithms:

1. ${\log }_{a}a=1$
2. ${\log }_a\left(x.y\right)={\log }_{a}x+{\log }_{a}y$
3. ${\log }_a\left(\frac{x}{y}\right)={\log }_{a}x-{\log }_{a}y$
4. ${\log }_a\left(\frac{1}{x}\right)=-{\log }_{a}x$
5. $\mathrm{l}\mathrm{o}{\mathrm{g}}_a{x}^p=p\mathrm{l}\mathrm{o}{\mathrm{g}}_{a}x$
6. $lo{g}_a\left(x\right)=\frac{lo{g}_b\left(x\right)}{lo{g}_b\left(a\right)}$

Calculation:

log 6 = log (2 x 3)

From 2nd property of logarithms

log 6 = log 2 + log 3

log 6 = x + y

Concept:

Chain rule: $\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}[\mathrm{f}(\mathrm{g}(\mathrm{x}))]={\mathrm{f}}^{\prime }(\mathrm{g}(\mathrm{x})){\mathrm{g}}^{\prime }(\mathrm{x})$

If y = log x, then f'(x) = 1/x

And if y = xⁿ, then f'(x) = nx^{n - 1}

Calculation:

f(x) = log x², x > 1

Differentiating with respect to x, we get

⇒ f'(x) = ($\frac{1}{{\mathrm{x}}^2}$)$\frac{\mathrm{d}{\mathrm{x}}^2}{\mathrm{d}\mathrm{x}}$ 

⇒ f'(x) = $\frac{2\mathrm{x}}{{\mathrm{x}}^2}$

⇒ f'(x) = $\frac{2}{\mathrm{x}}$

Concept:

Suppose that we have two functions f(x) and g(x) and they are both differentiable.

• Chain Rule: $\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\left[\mathrm{f}\left(\mathrm{g}\left(\mathrm{x}\right)\right)\right]={\mathrm{f}}^{\prime }\left(\mathrm{g}\left(\mathrm{x}\right)\right){\mathrm{g}}^{\prime }\left(\mathrm{x}\right)$
• Product Rule: $\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\left[\mathrm{f}\left(\mathrm{x}\right)\mathrm{g}\left(\mathrm{x}\right)\right]={\mathrm{f}}^{\prime }\left(\mathrm{x}\right)\mathrm{g}\left(\mathrm{x}\right)+\mathrm{f}\left(\mathrm{x}\right){\mathrm{g}}^{\prime }\left(\mathrm{x}\right)$
• log ab = b log a 

Calculation:-  

${\mathrm{x}}^{\mathrm{m}}{\mathrm{y}}^{\mathrm{n}}=\mathrm{x}\mathrm{y}{\mathrm{a}}^{\mathrm{m}+\mathrm{n}}$ , where a is constant 

Taking log both sides, we get

⇒ $\log ({\mathrm{x}}^{\mathrm{m}}{\mathrm{y}}^{\mathrm{n}})=\log (\mathrm{x}\mathrm{y}{\mathrm{a}}^{\mathrm{m}+\mathrm{n}})$

⇒ log x^m + log yⁿ = log x + log y + log a^m+n

⇒ m log x + n log y = log x + log y + (m + n) log a 

Differentiate both sides w.r.t  x, we get

⇒ $\frac{\mathrm{m}}{\mathrm{x}}+\frac{\mathrm{n}}{\mathrm{y}}\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=\frac{1}{\mathrm{x}}+\frac{1}{\mathrm{y}}\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}+0$   [∵  $\frac{\mathrm{d}\text{constant}}{\mathrm{d}\mathrm{x}}=0$ ]  

⇒ $\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}[\frac{\mathrm{n}}{\mathrm{y}}-\frac{1}{\mathrm{y}}]=[\frac{1}{\mathrm{x}}-\frac{\mathrm{m}}{\mathrm{x}}]$ 

⇒  $\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=\frac{-(\mathrm{m}-1)\mathrm{y}}{(\mathrm{n}-1)\mathrm{x}}$ . 

The correct option is 3. 

Concept:

Chain rule: 

$\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}[\mathrm{f}(\mathrm{g}(\mathrm{x}))]={\mathrm{f}}^{\prime }(\mathrm{g}(\mathrm{x})){\mathrm{g}}^{\prime }(\mathrm{x})$

If y = log x, then f'(x) = 1/x

And if y = √x, then f'(x) = $\frac{1}{2\sqrt{\mathrm{x}}}$

Calculation:

f(x) = log √x

Differentiating with respect to x, we get

⇒ f'(x) = ($\frac{1}{\sqrt{\mathrm{x}}}$)$\frac{\mathrm{d}\sqrt{\mathrm{x}}}{\mathrm{d}\mathrm{x}}$ 

⇒ f'(x) = $\frac{1}{\sqrt{\mathrm{x}}}$⋅ $\frac{1}{2\sqrt{\mathrm{x}}}$

⇒ f'(x) = $\frac{1}{2\mathrm{x}}$

Concept:

Chain rule: $\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}[\mathrm{f}(\mathrm{g}(\mathrm{x}))]={\mathrm{f}}^{\prime }(\mathrm{g}(\mathrm{x})){\mathrm{g}}^{\prime }(\mathrm{x})$

If y = log x, then f'(x) = 1/x

And if y = sin x, then f'(x) = cos x

Calculation:

f(x) = log (sin x)

Differentiating with respect to x, we get

⇒ f'(x) = ($\frac{1}{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{x}}$)$\frac{\mathrm{d}\sin \mathrm{x}}{\mathrm{d}\mathrm{x}}$ 

⇒ f'(x) = $\frac{\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{x}}{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{x}}$

⇒ f'(x) = cot x

Concept:

Chain rule: 

$\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}[\mathrm{f}(\mathrm{g}(\mathrm{x}))]={\mathrm{f}}^{\prime }(\mathrm{g}(\mathrm{x})){\mathrm{g}}^{\prime }(\mathrm{x})$

Calculation:

Here, -log (log x), x > 1

Let, log x = y 

Differentiating with respect to x, we get

⇒ dy/dx = 1/x        ....(1)

Now, -log (log x) = -log y

$\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}(-\log \mathrm{y})=-\frac{1}{\mathrm{y}}\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}$

= -1 / (x log x)    ....(from (1))

Hence, option (1) is correct. 

Given the values:

log 2 = 0.2614

log 3 = 0.3521

Formula Used:

Using the logarithm property:

Calculation:

We can write as 6 = 2 x 3

Using the property:

log 6 = log 2 + log 3

Substituting the given values:

⇒ log 6 = 0.6135

Hence the value log 6 is 0.6135.

Therefore, the correct option is (2)

Concept:

Formula:

log mn = n log m

$\frac{\mathrm{d}(\mathrm{u}\mathrm{v})}{\mathrm{d}\mathrm{x}}=\mathrm{v}\frac{\mathrm{d}\mathrm{u}}{\mathrm{d}\mathrm{x}}+\mathrm{u}\frac{\mathrm{d}\mathrm{v}}{\mathrm{d}\mathrm{x}}$

$\frac{\mathrm{d}\log \mathrm{x}}{\mathrm{d}\mathrm{x}}=\frac{1}{\mathrm{x}}$

$\frac{\mathrm{d}\mathrm{x}}{\mathrm{d}\mathrm{x}}=1$

Calculation:

Let y = x^log x

Taking log both sides, we get

⇒ log y = x^{log x}

⇒ log y = log x log  x            (∵ log mn = n log m)

Differentiating with respect to x, we get

⇒ $\frac{1}{\mathrm{y}}\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=\log \mathrm{x}\frac{\mathrm{d}\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{x}}{\mathrm{d}\mathrm{x}}+\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{x}\frac{\mathrm{d}\log \mathrm{x}}{\mathrm{d}\mathrm{x}}$

⇒ $\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=\mathrm{y}(\log \mathrm{x}\times \frac{1}{\mathrm{x}}+\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{x}\times \frac{1}{\mathrm{x}})$

⇒ $\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=\frac{{\mathrm{x}}^{\log \mathrm{x}}}{\mathrm{x}}(2\log \mathrm{x})$

⇒ $\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}={\mathrm{x}}^{\log \mathrm{x}-1}(\log {\mathrm{x}}^2)$