Logarithmic Function MCQ Quiz - Objective Question with Answer for Logarithmic Function - Download Free PDF
Last updated on Apr 22, 2025
Latest Logarithmic Function MCQ Objective Questions
Logarithmic Function Question 1:
If y = In(emx + e-mx), then what is the value of \(\rm \frac{dy}{dx}\) at x = 0 ?
Answer (Detailed Solution Below)
Logarithmic Function Question 1 Detailed Solution
Concept:
If y = ln f(x) then \(\frac{dy}{dx}=\frac{f'(x)}{f(x)}\)
Calculation:
Given function is y = ln(emx + 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.
Logarithmic Function Question 2:
If \(\rm y = \log \left( \frac{1 + \sqrt x}{1 - \sqrt x} \right)\) then \(\rm \frac{dy}{dx} \) is equal to
Answer (Detailed Solution Below)
Logarithmic Function Question 2 Detailed Solution
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 xn-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).
Logarithmic Function Question 3:
If x2y - 2x + y = 0; |x| < 1, then \(\rm \left[y+\frac{y^3}{3}+\frac{y^5}{5}+...\right]/\left[x+\frac{x^3}{3}+\frac{x^5}{5}+....\right]\) is equal to-
Answer (Detailed Solution Below)
Logarithmic Function Question 3 Detailed Solution
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.
Logarithmic Function Question 4:
If \(\log_5 \log_8 (x^2 - 1) = 0\), then a possible value of \(x\) is
Answer (Detailed Solution Below)
Logarithmic Function Question 4 Detailed Solution
To solve the equation \(\log_5 \log_8 (x^2 - 1) = 0\), follow these steps:
- Remove the outer logarithm:
\(\log_5 \log_8 (x^2 - 1) = 0 \implies \log_8 (x^2 - 1) = 5^0 = 1\) - Remove the inner logarithm:
\(\log_8 (x^2 - 1) = 1 \implies x^2 - 1 = 8^1 = 8\) - Solve for \(x^2\):
\(x^2 - 1 = 8 \implies x^2 = 9\) - Solve for \(x\):
\(x = \pm 3\)
Therefore, the possible values of x are \(3\) and \(-3\).
Logarithmic Function Question 5:
The inequality \( \log_a f(x) < \log_a g(x) \) implies that
Answer (Detailed Solution Below)
f(x) > g(x) > 0 for 0 < a < 1 and g(x) > f(x) > 0 for a > 1
Logarithmic Function Question 5 Detailed Solution
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.
Top Logarithmic Function MCQ Objective Questions
If p > q >1 and 0 < a < 1 then which of the following is correct option?
Answer (Detailed Solution Below)
Logarithmic Function Question 6 Detailed Solution
Download Solution PDFExplanation:
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 = logax is a decreasing function for base a ∈ (0, 1). So, while operating loga 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 loga(p – a) and loga(q – a).
Operating log function on both sides,
⇒ loga(p - a) < loga(q - a)
⇒ Option 3 is correct.
Find out the value of x if logx 4 + logx 16 + logx 64 = 12
Answer (Detailed Solution Below)
Logarithmic Function Question 7 Detailed Solution
Download Solution PDFGiven:
logx 4 + logx 16 + logx 64 = 12
Formula Used:
If logx y = a, then xa = y
logx ab = b logx a
Calculation:
logx 4 + logx 16 + logx 64 = 12
⇒ logx 22 + logx 24 + logx 26 = 12
⇒ 2 logx 2 + 4 logx 2 + 6 logx 2 = 12
⇒ 12 logx 2 = 12
⇒ logx 2 = 1
⇒ 2 = x1
∴ x = 2
If log (x + 2) + log (x − 2) = log 5 then the value of x will be
Answer (Detailed Solution Below)
Logarithmic Function Question 8 Detailed Solution
Download Solution PDFConcept:
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
A radioactive substance decays at a rate proportional to the amount of substance present. If half of the substance decays in 100 years, then what is the decay constant (proportionality constant) ?
Answer (Detailed Solution Below)
Logarithmic Function Question 9 Detailed Solution
Download Solution PDFConcept
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 t1/2 = 100 years,
⇒ 100 = \(\displaystyle \frac{log_e2}{k}\)
⇒ k = \(\displaystyle \frac{log_e2}{100}\)
∴ Decay constant is \(\displaystyle \frac{log_e2}{100}\).
The relation between half life (T) and decay constant (λ) is:
Answer (Detailed Solution Below)
Logarithmic Function Question 10 Detailed Solution
Download Solution PDFCONCEPT:
- 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 = loge 2
- So the correct answer will be option 3.
If x = 27 and y = log3 4, then xy equals
Answer (Detailed Solution Below)
Logarithmic Function Question 11 Detailed Solution
Download Solution PDFConcept:
- If m and n are positive real numbers then, loga(mn) = nlogam.
- If a is a positive real number and m is a positive rational number then, \({a^{{{\log }_a}m}} = m\).
Calculation:
\({x^y} = {27^{{{\log }_3}4}} = {\left( {{3^3}} \right)^{{{\log }_3}4}}{\rm{ = }}{3^{3{{\log }_3}4}} \\= {3^{{{\log }_3}{4^3}}} = {4^3} = 64\)
If a and b are positive real numbers other than unity, then the least value of |logb a + loga b|, is
Answer (Detailed Solution Below)
Logarithmic Function Question 12 Detailed Solution
Download Solution PDFConcept:
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
- G2 = AH
- 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, logb 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, \(\rm \left| {{{\log }_b}a + {\rm{ }}\frac{1}{{{{\log }_b}a}}} \right| \ge 2\)
\(\frac{1}{{{{\log }_2}x}} + \frac{1}{{{{\log }_3}x}} + \frac{1}{{{{\log }_4}x}} + \ldots .. + \frac{1}{{{{\log }_{50}}x}},x \ne 1\) is equal to
Answer (Detailed Solution Below)
Logarithmic Function Question 13 Detailed Solution
Download Solution PDFConcept:
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:
- \({\log _a}a = 1\)
- \({\log _a}\left( {x.y} \right) = {\log _a}x + {\log _a}y\)
- \({\log _a}\left( {\frac{x}{y}} \right) = {\log _a}x - {\log _a}y\)
- \({\log _a}\left( {\frac{1}{x}} \right) = - {\log _a}x\)
- \({\rm{lo}}{{\rm{g}}_a}{x^p} = p{\rm{lo}}{{\rm{g}}_a}x\)
- \(lo{g_a}\left( x \right) = \frac{{lo{g_b}\left( x \right)}}{{lo{g_b}\left( a \right)}}\)
The value of Log327 is?
Answer (Detailed Solution Below)
Logarithmic Function Question 14 Detailed Solution
Download Solution PDFGiven:
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.
∫dx/x = log |(x)| is not possible when:
Answer (Detailed Solution Below)
Logarithmic Function Question 15 Detailed Solution
Download Solution PDFExplanation:
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).