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3 years ago

Solve log x = 2 by changing it to exponential form.

Mathematics

2 answers:

LiRa [457]3 years ago

7 0

Answer:

Step-by-step explanation:

Convert to Exponential Form log of x=-2. log(x)=−2 log ( x ) = - 2. For logarithmic equations, logb(x)=y log b ( x ) = y is equivalent to by=x b y = x such that x>0 x

Veronika [31]3 years ago

4 0

Answer:

Option B

Step-by-step explanation:

log x=2

x = 10^2

Therefore, the exponential form is the one in option B

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Find the equation of the straight line passing through the point (3, 5)

Ne4ueva [31]

Answer:

y = -1/3x + 6

Step-by-step explanation:

the slope is -1/3 because you find the the reciprocal and switch the sign. To find the y intercept you plug in the points to the equation.

5 = -1/3(3) + b

5 = -1 + b

6 = b

5 0

3 years ago

Find the area of the sector with a central angle of 120° and a radius of 8 inches. Leave in terms of π

soldier1979 [14.2K]

Centralangle/360 times area of circle=sector area

120/360 times pi8²=
(1/3)(64pi)=64pi/3 square inches

8 0

3 years ago

Read 2 more answers

What is 3.4749 rounded to the nearest hundred

Rus_ich [418]

Answer:

3.47

Step-by-step explanation:

since the third decimal is less than 5 then it doesn't carry over to the 7 thus it remains 7.

5 0

3 years ago

In which quadrant does θ lie if the following statements are true: cos⁡θ>0 and sin⁡θ>0

Papessa [141]

Answer:

First quadrant

Step-by-step explanation:

cosθ > 0 in first/ fourth quadrants

sinθ > 0 in the first/ second quadrants

sinθ and cosθ are both > 0 in the first quadrant

Then θ lies in the first quadrant

6 0

3 years ago

Consider the differential equation: xy′(x2+7)y=cos(x)+e3xy. Put the differential equation into the form: y′+p(x)y=g(x), determin

icang [17]

Answer:

Linear and non-homogeneous.

Step-by-step explanation:

We are given that

$\frac{xy'}{(x^2+7)y}=cosx+\frac{e^{3x}}{y}$

We have to convert into y'+P(x)y=g(x) and determine P(x) and g(x).

We have also find type of differential equation.

$y'=\frac{(x^2+7)y}{x}(cosx+\frac{e^{3x}}{y}}$

$y'=\frac{(x^2+7)cosx}{x}y+\frac{(x^2+7)e^{3x}}{x}$

$y'-\frac{cosx(x^2+7)}{x}y=\frac{e^{3x}(x^2+7)}{x}$

It is linear differential equation because this equation is of the form

y'+P(x)y=g(x)

Compare it with first order first degree linear differential equation

$y'+P(x)y=g(x)$

$P(x)=-\frac{cosx (x^2+7)}{x},g(x)=\frac{e^{3x}(x^2+7)}{x}$

$\frac{dy}{dx}=\frac{(x^2+7)(ycosx+e^{3x})}{x}$

Homogeneous equation

$\frac{dy}{dx}=\frac{f(x,y)}{g(x,y)}$

Degree of f and g are same.

$f(x,y)=(x^2+7)(ycosx+e^{3x}),g(x,y)=x$

Degree of f and g are not same .

Therefore, it is non- homogeneous .

Linear and non-homogeneous.

3 0

3 years ago