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

Concept:

If y = ln f(x) then \(\frac{dy}{dx}=\frac{f'(x)}{f(x)}\)
Calculation:

Given function is y = ln(e^mx + me^-mx)

Differentiating, we get

\(\frac{dy}{dx}=\frac{1}{e^{mx}+e^{-mx}}(me^{mx}-me^{-mx})\)

At x = 0, we have

\(\frac{dy}{dx}=\frac{1}{e^{0}+e^{0}}(me^{m0}-me^{-m0})\)

⇒ \(\frac{dy}{dx}=0\)

∴ For y = ln(emx + me-mx), \(\frac{dy}{dx}=0\) at x = 0.

Concept:

Chain Rule: 

• If y = f(g(x)), then dy/dx = f '(g(x) × g'(x)
• If y = p(x)/q(x), then dy/dx = \({(p'(x)q(x) - p(x) q'(x)) \over (q(x))^2}\)

Formula used:

• (log x)' = 1/x
• (xⁿ)' = n x^n-1

Calculation:

If  \(\rm y = \log \left( \frac{1 + \sqrt x}{1 - \sqrt x} \right)\),

then, \({dy\over dx} = {1\over \left( \frac{1 + \sqrt x}{1 - \sqrt x} \right)} \times {d\over dx}\left( \frac{1 + \sqrt x}{1 - \sqrt x} \right)\)

⇒ \({dy\over dx} = { \left( \frac{1 - \sqrt x}{1 + \sqrt x} \right)} \times \frac{({1\over 2\sqrt x})(1 - \sqrt x) - (1+\sqrt x)({-1\over 2\sqrt x})}{(1 - \sqrt x)^2} \)

⇒ \({dy\over dx} = { \left( \frac{1 - \sqrt x}{1 + \sqrt x} \right)} \times ({1\over 2\sqrt x})\frac{(1 - \sqrt x) + (1+\sqrt x)}{(1 - \sqrt x)^2} \)

⇒ \({dy\over dx} = { \left( \frac{1 - \sqrt x}{1 + \sqrt x} \right)} \times ({1\over 2\sqrt x})\frac{ 2}{(1 - \sqrt x)^2} \)

⇒ \({dy\over dx} = { \left( \frac{1 - \sqrt x}{1 + \sqrt x} \right)} \times ({1\over \sqrt x})\frac{ 1}{(1 - \sqrt x)^2} \)

⇒ \({dy\over dx} = { \left( \frac{1}{\sqrt x(1 + \sqrt x)(1 - \sqrt x)} \right)} \)

⇒ \({dy\over dx} = { \left( \frac{1}{\sqrt x (1 - x)} \right)} \)

∴ The correct answer is option (4).

Explanation:

We know that

log(1 + x) = x + \({x^2\over 2}+{x^3\over 3}+{x^5\over 5}+...\)

and 

log(1 - x) = -x + \({x^2\over 2}-{x^3\over 3}+{x^5\over 5}-...\)

Subtracting them we get

\(x+\frac{x^3}{3}+\frac{x^5}{5}+....=\frac12\log(\frac{1+x}{1-x})\)

Therefore

\(\rm \left[y+\frac{y^3}{3}+\frac{y^5}{5}+...\right]/\left[x+\frac{x^3}{3}+\frac{x^5}{5}+....\right]\) = \(\frac12\log(\frac{1+y}{1-y})\over \frac12\log(\frac{1+x}{1-x})\)

\(\rm \left[y+\frac{y^3}{3}+\frac{y^5}{5}+...\right]/\left[x+\frac{x^3}{3}+\frac{x^5}{5}+....\right]\) = \(\log(\frac{1+y}{1-y})\over \log(\frac{1+x}{1-x})\)...(i)

Now, given x2y - 2x + y = 0

i.e., (x² + 1)y = 2x ⇒ y = \(2x\over 1+x^2\)

Hence 

\(\log(\frac{1+y}{1-y})=\log(\frac{1+{2x\over 1+x^2}}{1-{2x\over 1+x^2}})\)

             = \(\log(\frac{{(1+x)^2\over 1+x^2}}{{(1-x)^2\over 1+x^2}})\)

            = \(\log(\frac{1+x}{1-x})^2\)

           = \(2\log(\frac{1+x}{1-x})\)

Putting this value in (i) we get

\(\rm \left[y+\frac{y^3}{3}+\frac{y^5}{5}+...\right]/\left[x+\frac{x^3}{3}+\frac{x^5}{5}+....\right]\) = \(2\log(\frac{1+x}{1-x})\over \log(\frac{1+x}{1-x})\) = 2

Option (2) is true.

To solve the equation \(\log_5 \log_8 (x^2 - 1) = 0\), follow these steps:

1. Remove the outer logarithm:
   \(\log_5 \log_8 (x^2 - 1) = 0 \implies \log_8 (x^2 - 1) = 5^0 = 1\)
2. Remove the inner logarithm:
   \(\log_8 (x^2 - 1) = 1 \implies x^2 - 1 = 8^1 = 8\)
3. Solve for \(x^2\):
   \(x^2 - 1 = 8 \implies x^2 = 9\)
4. Solve for \(x\):
   \(x = \pm 3\)

Therefore, the possible values of x are \(3\) and \(-3\).

the answer is f(x) > g(x) > 0 for 0 < a < 1 and g(x) > f(x) > 0 for a > 1 (using properties of logarithmic inequalities)

Option 4.

Explanation:

For a log function with base a as 0 < a < 1, the graph of the function can be drawn as bellow

The above graph is valid for,

x > 0 and 0 < a < 1

Since the function y = log_ax is a decreasing function for base a ∈ (0, 1). So, while operating log_a on a given inequality, the sign of the inequality will change.

Calculation:

Given:

p > q > 1

As, 0 < a < 1

⇒ (p – a) > (q – a) > 0

This is satisfying the condition for the domain of log_a(p – a) and loga(q – a).

Operating log function on both sides,

⇒ log_a(p - a) < log_a(q - a)

⇒ Option 3 is correct.

Given:

log_x 4 + log_x 16 + log_x 64 = 12

Formula Used:

If log_x y = a, then x^a = y

log_x a^b = b log_x a

Calculation:

log_x 4 + log_x 16 + log_x 64 = 12

⇒ log_x 2² + log_x 2⁴ + log_x 2⁶ = 12

⇒ 2 log_x 2 + 4 log_x 2 + 6 log_x 2 = 12

⇒ 12 log_x 2 = 12

⇒ log_x 2 = 1

⇒ 2 = x¹

∴ x = 2

Concept:

Logarithm properties

Product rule: The log of a product equals the sum of two logs.

\({\log _{\rm{a}}}\left( {{\rm{mn}}} \right) = {\rm{\;}}{\log _{\rm{a}}}{\rm{m}} + {\rm{\;}}{\log _{\rm{a}}}{\rm{n}}\)

Quotient rule: The log of a quotient equals the difference of two logs.

\({\log _{\rm{a}}}\frac{{\rm{m}}}{{\rm{n}}} = {\rm{\;}}{\log _{\rm{a}}}{\rm{m}} - {\rm{\;}}{\log _{\rm{a}}}{\rm{n}}\)

Calculation:

Given:

log (x + 2) + log (x − 2) = log 5

⇒ log [(x + 2) (x – 2)] = log 5                (∵ \({\log _{\rm{a}}}\left( {{\rm{mn}}} \right) = {\rm{\;}}{\log _{\rm{a}}}{\rm{m}} + {\rm{\;}}{\log _{\rm{a}}}{\rm{n}}\))

⇒ log (x ² – 4) = log 5

⇒ x² – 4 = 5

⇒ x² = 9

∴ x = ± 3

Here x ≠ -3 is not possible because (x – 2) should be greater than zero.

∴ x = 3

Concept

Half-life:  \(\displaystyle t_{\frac{1}{2}}=\frac{log_e2}{k}\)

Where k is the decay constant.

Calculation:

We know that  \(\displaystyle t_{\frac{1}{2}}=\frac{log_e2}{k}\)

Putting the value of t_1/2 = 100 years,

⇒ 100 = \(\displaystyle \frac{log_e2}{k}\)

⇒ k = \(\displaystyle \frac{log_e2}{100}\)

∴ Decay constant is \(\displaystyle \frac{log_e2}{100}\).

CONCEPT:

• As per radioactive decay law, the total number of nuclei of radioactive compounds after radioactive decay in the sample is given by given equation

\(N=N_0 e^{-λ t}\)

where N is the number of nuclei of radioactive compounds after radioactive decay, N₀ is the number of nuclei of radioactive compounds initially, λ is the decay constant and t is the time of radioactive decay.

• Half-Life: Half-life is the time required for a radioactive substance to reduce to its half initial value.

CALCULATION:

• From the definition of half-life, radioactive decay in time T (half-life) will be half of its initial value

N = N₀ / 2

so at time t = T, no of nuclei is N = N₀ / 2

where N₀ is the initial number of nuclei.

From radioactive decay law:

\(⇒ N=N_0 e^{-λ t}\)

\(⇒ \frac{N_0}{2}=N_0 e^{-λ T}\)

\(⇒ \frac{1}{2}= e^{-λ T}\)

\(⇒ e^{λ T} = 2\)

Taking log on both the sides

⇒ λ T = log_e 2

• So the correct answer will be option 3.

Concept:

• If m and n are positive real numbers then, log_a(mⁿ) = nlog_am.
• If a is a positive real number and m is a positive rational number then, \({a^{{{\log }_a}m}} = m\).     

Calculation:

\({x^y} = {27^{{{\log }_3}4}} = {\left( {{3^3}} \right)^{{{\log }_3}4}}{\rm{ = }}{3^{3{{\log }_3}4}} \\= {3^{{{\log }_3}{4^3}}} = {4^3} = 64\)

Concept:

AM, GM, HM Formulas

If A is the arithmetic mean of numbers a and b and is given by ⇔ \({\rm{A}} = \frac{{{\rm{a\;}} + {\rm{\;b}}}}{2}\)

If G is the geometric mean of the numbers a and b and is given by ⇔ \({\rm{G}} = \sqrt {{\rm{ab}}} \)

If H is the Harmonic mean of numbers a and b and is given by ⇔ \({\rm{H}} = \frac{{2{\rm{ab}}}}{{{\rm{a}}\ +\ {\rm{b}}}}\)

Relation between AM, GM and HM

1. G2 = AH
2. AM  ≥  GM  ≥  HM

Change of base rule

\({\log _m}n = \frac{{{{\log }_a}n}}{{{{\log }_a}m}}\)

Calculation:

\(\rm \left| {{{\log }_b}a + {\rm{ lo}}{{\rm{g}}_a}b} \right| = \left| {{{\log }_b}a + {\rm{ }}\frac{1}{{{{\log }_b}a}}} \right|\)

Let, log_b a = x

Then, \(\rm \left| {{{\log }_b}a + {\rm{ }}\frac{1}{{{{\log }_b}a}}} \right| = \left| {x + {\rm{ }}\frac{1}{x}} \right|\)

We know that,

AM ≥ GM

⇒ \(\rm \frac { x \ +\ \frac 1 x }2 \geq \sqrt {x \times \frac 1 x}\)

⇒ \(\rm x + \frac 1 x \geq 2\)

⇒ \(\rm \left| x + \frac 1 x \right| \geq 2\)

So,  \(\rm \left| {{{\log }_b}a + {\rm{ }}\frac{1}{{{{\log }_b}a}}} \right| \ge 2\)

Concept:

We know that,

\({\log _x}y = \frac{1}{{{{\log }_y}x}}\) ---(1)

Calculation:

Given:

\(\frac{1}{{{{\log }_2}x}} + \frac{1}{{{{\log }_3}x}} + \frac{1}{{{{\log }_4}x}} + \ldots .. + \frac{1}{{{{\log }_{50}}x}},x \ne 1\)

By using above property, It can be written as:

\({\log _x}2 + {\log _x}3 + \ldots + {\log _x}50\;\;\;\;\;\;\;\;\;\;\)

\({\log _x}\left[ {2 \times 3 \times 4 \times 5 \times \ldots \times 50} \right]\)

\({\log _x}50! \Rightarrow \frac{1}{{{{\log }_{50!}}x}}\)

Important Points

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. \({\rm{lo}}{{\rm{g}}_a}{x^p} = p{\rm{lo}}{{\rm{g}}_a}x\)
6. \(lo{g_a}\left( x \right) = \frac{{lo{g_b}\left( x \right)}}{{lo{g_b}\left( a \right)}}\)

Given:

We have a logarithm expression 'Log327'

Concept:

Logarithm

Formula Used:

Logab = x 

ax = b

Calculation:

Log327 = n

⇒ 3n = 27

⇒ 3n = 33

⇒ n = 3
∴ The value of Log327 is 3.

Explanation:

Consider the function \(f(x)=\frac{1}{x}\)

with the intention of examining the function's continuity for all real numbers, we will find that the function is discontinuous at x = 0, and hence the function is not integrable at x = 0.

However, if we review it within the region of all positive real numbers or all negative real numbers, then the function is continuous, and also integrable in this region.

This is the consolation we needed that the function is integrable for all positive real numbers (or negative numbers).

Moreover, the integration of the function \(\frac{1}{x}\) is ln|(x)|, which is not defined for x = 0.

So this further confirms the fact that \(\frac{1}{x}\) is not integrable only for x = 0 (since the integration of \(\frac{1}{x}\) is not defined for x = 0).