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.

log_{b}(*xy*) = log_{b}*x* + log_{b}*y*.

log_{b}(*x/y*) = log_{b}*x* - log_{b}*y*.

log_{b}(*x ^{n}*) =

*n*log

_{b}

*x*.

log

_{b}

*x*= log

_{a}

*x*/ log

_{a}

*b*.

### 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.