All categories

3 years ago

Which expression is equivalent to log5(X/4)^2

Mathematics

2 answers:

saw5 [17]3 years ago

4 0

Answer: Applying properties of logarithm, the expression equivalent to

log5 (x/4)^2 is: 2 (log5 x - log5 4)

Solution:

log5 (x/4)^2

Using loga b^c = c loga b; with a=5, b=(x/4), and c=2

log5 (x/4)^2 = 2 log5 (x/4)

Using loga (b/c) = loga b - loga c; with a=5, b=x, and c=4

log5 (x/4)^2 = 2 (log5 x - log5 4)

zzz [600]3 years ago

3 0

2log(base 5)x - 2log(base 5)4

You might be interested in

A survey is made to determine the number of households having electric appliances in a certain city. It is found that 75% have r

Mashcka [7]

Answer:

The probability that a household has at least one of these appliances is 0.95

Step-by-step explanation:

Percentage of households having radios P(R) = 75% = 0.75

Percentage of households having electric irons P(I) = 65% = 0.65

Percentage of households having electric toasters P(T) = 55% = 0.55

Percentage of household having iron and radio P(I∩R) = 50% = 0.5

Percentage of household having radios and toasters P(R∩T) = 40% = 0.40

Percentage of household having iron and toasters P(I∩T) = 30% = 0.30

Percentage of household having all three P(I∩R∩T) = 20% = 0.20

Probability of households having at least one of the appliance can be calculated using the rule:

P(at least one of the three) = P(R) +P(I) + P(T) - P(I∩R) - P(R∩T) - P(I∩T) + P(I∩R∩T)

P(at least one of the three)=0.75 + 0.65 + 0.55 - 0.50 - 0.40 - 0.30 + 0.20 P(at least one of the three) = 0.95

The probability that a household has at least one of these appliances is 0.95

3 0

3 years ago

ANSWER QUICK!!!<br><br> PLEASE INCLUDE WORKKKK

alexgriva [62]

Answer:

725

Step-by-step explanation:

700-7125 sksksksksks

7 0

3 years ago

Maths functions question!!

Marina86 [1]

Answer:

5) DE = 7 units and DF = 4 units

6) ST = 8 units

$\textsf{7)} \quad \sf OM=\dfrac{3}{2}\:units$

8) x ≤ -3 and x ≥ 3

Step-by-step explanation:

<u>Information from Parts 1-4:</u>

brainly.com/question/28193969

$f(x)=-x+3$

$g(x)=x^2-9$

A = (3, 0) and C = (-3, 0)

Points A and D are the <u>points of intersection</u> of the two functions.

To find the x-values of the points of intersection, equate the two functions and solve for x:

$\implies g(x)=f(x)$

$\implies x^2-9=-x+3$

$\implies x^2+x-12=0$

$\implies x^2+4x-3x-12=0$

$\implies x(x+4)-3(x+4)=0$

$\implies (x-3)(x+4)=0$

Apply the zero-product property:

$\implies (x-3)= \implies x=3$

$\implies (x+4)=0 \implies x=-4$

From inspection of the graph, we can see that the x-value of point D is <u>negative</u>, therefore the x-value of point D is x = -4.

To find the y-value of point D, substitute the found value of x into one of the functions:

$\implies f(-4)=-(-4)=7$

Therefore, D = (-4, 7).

The length of DE is the difference between the y-value of D and the x-axis:

⇒ DE = 7 units

The length of DF is the difference between the x-value of D and the x-axis:

⇒ DF = 4 units

To find point S, substitute the x-value of point T into function g(x):

$\implies g(4)=(4)^2-9=7$

Therefore, S = (4, 7).

The length ST is the difference between the y-values of points S and T:

$\implies ST=y_S-y_T=7-(-1)=8$

Therefore, ST = 8 units.

The given length of QR (⁴⁵/₄) is the difference between the functions at the same value of x. To find the x-value of points Q and R (and therefore the x-value of point M), subtract g(x) from f(x) and equate to QR, then solve for x:

$\implies f(x)-g(x)=QR$

$\implies -x+3-(x^2-9)=\dfrac{45}{4}$

$\implies -x+3-x^2+9=\dfrac{45}{4}$

$\implies -x^2-x+\dfrac{3}{4}=0$

$\implies -4\left(-x^2-x+\dfrac{3}{4}\right)=-4(0)$

$\implies 4x^2+4x-3=0$

$\implies 4x^2+6x-2x-3=0$

$\implies 2x(2x+3)-1(2x+3)=0$

$\implies (2x-1)(2x+3)=0$

Apply the zero-product property:

$\implies (2x-1)=0 \implies x=\dfrac{1}{2}$

$\implies (2x+3)=0 \implies x=-\dfrac{3}{2}$

As the x-value of points M, Q and P is negative, x = -³/₂.

Length OM is the difference between the x-values of points M and the origin O:

$\implies x_O-x_m=o-(-\frac{3}{2})=\dfrac{3}{2}$

Therefore, OM = ³/₂ units.

The values of x for which g(x) ≥ 0 are the values of x when the parabola is above the x-axis.

Therefore, g(x) ≥ 0 when x ≤ -3 and x ≥ 3.

8 0

1 year ago

Read 2 more answers

Does anyone know a rule that work for all of these

Oxana [17]

Divide each term in the y column by 2 to go from this list {8, 2, 0, 2, 8} to this list {4, 1, 0, 1, 4}

This list {4, 1, 0, 1, 4} is a bunch of perfect squares so it suggests that y = x^2

However we must double the values to get back to the original list. So the rule is y = 2*x^2

3 0

3 years ago

Which conditions describe a rectangle?

N76 [4]

Answer:

The answer is B 1, 2, and 3

5 0

3 years ago

Read 2 more answers