Question
Given that 
and 
then value of 
equals –
The correct answer is:
Here we have to find the value of if 
and .
{ We know that 
}
Also we can write as 
And ,
Also we can write as 
.
If a, m and n are positive integers and a ≠ 1, then;
Thus, the log of two numbers m and n, with base ‘a’ is equal to the sum of log m and log n with the same base ‘a’.
These four basic properties all follow directly from the fact that logs are exponents.
logb(xy) = logbx + logby.
logb(x/y) = logbx - logby.
logb(xn) = n logbx.
logbx = logax / logab.
Related Questions to study
Statement 1:Change in colour of acidic solution of potassium dichromate by breath is used to test drunk drivers.
Statement 2:Change in colour is due to the complexation of alcohol with potassium dichromate.
Statement 1:Change in colour of acidic solution of potassium dichromate by breath is used to test drunk drivers.
Statement 2:Change in colour is due to the complexation of alcohol with potassium dichromate.
Total number of solutions of sin{x} = cos{x}, where {.} denotes the fractional part, in [0, 2
] is equal to
In this question, we have to find the number of solutions. Here we have fractional part of x. It is {x} and {x}= x-[x]. The {x} is always belongs to 0 to 1.
Total number of solutions of sin{x} = cos{x}, where {.} denotes the fractional part, in [0, 2] is equal to
In this question, we have to find the number of solutions. Here we have fractional part of x. It is {x} and {x}= x-[x]. The {x} is always belongs to 0 to 1.
Statement 1:If a strong acid is added to a solution of potassium chromate it changes itscolour from yellow to orange.
Statement 2:The colour change is due to the oxidation of potassium chromate.
Statement 1:If a strong acid is added to a solution of potassium chromate it changes itscolour from yellow to orange.
Statement 2:The colour change is due to the oxidation of potassium chromate.
Total number of solutions of the equation 3x + 2 tan x =
in x 
[0, 2] is equal to
In this question, we use the graph of tanx . The intersection is the total number of solutions of this equation. The graph region is [ 0, 2π ].
Total number of solutions of the equation 3x + 2 tan x = in x [0, 2] is equal to
In this question, we use the graph of tanx . The intersection is the total number of solutions of this equation. The graph region is [ 0, 2π ].
The number of sigma bonds in 
is:
The number of sigma bonds in is:
Statement 1:Oxidation number of Ni in is zero.
Statement 2:Nickel is bonded to neutral ligand carbonyl.
Statement 1:Oxidation number of Ni in is zero.
Statement 2:Nickel is bonded to neutral ligand carbonyl.
Suppose equation is f(x) – g(x) = 0 of f(x) = g(x) = y say, then draw the graphs of y = f(x) and y = g(x). If graph of y = f(x) and y = g(x) cuts at one, two, three, ...., no points, then number of solutions are one, two, three, ...., zero respectively.
The number of solutions of sin x = 
is
In this question, we have drawn the graph. The number of intersections of both function fx and gx are the number solutions. Draw the graph carefully and find the intersection points.
Suppose equation is f(x) – g(x) = 0 of f(x) = g(x) = y say, then draw the graphs of y = f(x) and y = g(x). If graph of y = f(x) and y = g(x) cuts at one, two, three, ...., no points, then number of solutions are one, two, three, ...., zero respectively.
The number of solutions of sin x = is
In this question, we have drawn the graph. The number of intersections of both function fx and gx are the number solutions. Draw the graph carefully and find the intersection points.
The first noble gas compound obtained was:
The first noble gas compound obtained was:
Statement 1:The redox titrations in which liberated 
is used as oxidant are called as iodometric titrations
Statement 2:Addition of KI of 
liberates which is estimated against hypo solution.
Statement 1:The redox titrations in which liberated is used as oxidant are called as iodometric titrations
Statement 2:Addition of KI of liberates which is estimated against hypo solution.
molecule is completely changed into 
molecules at:
molecule is completely changed into molecules at:
Statement-I : If 
and 
then 
Statement-II : If sinA = sinB and cosA = cosB, then 
In this question, we have to find the statements are the correct or not and statement 2 is correct explanation or not, is same as assertion and reason. Here, Start solving first Statement and try to prove it. Then solve the Statement-II.
Statement-I : If and then
Statement-II : If sinA = sinB and cosA = cosB, then
In this question, we have to find the statements are the correct or not and statement 2 is correct explanation or not, is same as assertion and reason. Here, Start solving first Statement and try to prove it. Then solve the Statement-II.
Arsine is:
Arsine is:
Statement-I : The equation sin(cos x) = cos(sin x) does not possess real roots.
Statement-II : If sin x > 0, then 
Statement-I : The equation sin(cos x) = cos(sin x) does not possess real roots.
Statement-II : If sin x > 0, then
Statement-I : In (0, ), the number of solutions of the equation 
is two
Statement-II : 
is not defined at 
In this question, we have to find the statements are the correct or not and statement 2 is correct explanation or not, is same as assertion and reason. Here, it has 5 solutions, but tanθ &tan3θ are not defined at 
, 
, 
. respectively so it remains only 2.
Statement-I : In (0, ), the number of solutions of the equation is two
Statement-II : is not defined at
In this question, we have to find the statements are the correct or not and statement 2 is correct explanation or not, is same as assertion and reason. Here, it has 5 solutions, but tanθ &tan3θ are not defined at , , . respectively so it remains only 2.
Statement-I : If sin x + cos x = 
then 
Statement-II : AM ≥ GM
In this question, we have to find the statements are the correct or not and statement 2 is correct explanation or not, is same as assertion and reason. Here, Start solving first Statement and try to prove it. Then solve the Statement-II. Always, the AM–GM inequality states that AM ≥ GM.
Statement-I : If sin x + cos x = then
Statement-II : AM ≥ GM
In this question, we have to find the statements are the correct or not and statement 2 is correct explanation or not, is same as assertion and reason. Here, Start solving first Statement and try to prove it. Then solve the Statement-II. Always, the AM–GM inequality states that AM ≥ GM.
