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Tomtit [17]
3 years ago
15

What is the solution of logx 729=3?

Mathematics
2 answers:
shusha [124]3 years ago
3 0

Answer:

Solution

Step-by-step explanation:

Rewrite logx729= 3 is an exponential form using the definition of a logarithm. If x and b are positive real numbers and b≠1, then logb(x)= y is equivalent to b^y = x.

X^3 = 729

Take cube root on both side and we get

x= 3√729

now we firstly simplify the 3√729

Rewrite 729 as 9^3

x = 3√9^3

pull terms out from under the redical, assuming positive real numbers

x=9

verify each of the solution by substituting them into logx729=3 and solving.

x= 9

enot [183]3 years ago
3 0

For this case we have to define properties of logarithm that:

log_ {b} (a) = c

It is equivalent to:

b^c = a

So, given the following expression we have:

log_ {x} (729) = 3

We can rewrite it as:

x ^ 3 = 729

Applying cubic root to both sides of the equation we have:

x = \sqrt [3] {729}\\x = 9

ANswer:

x = 9

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29.)c-2.5

Step-by-step explanation:

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8 0
3 years ago
A. Complete the chart based on the initial conditions:
Nata [24]

Answer:

a.

Month\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Pop(whole \ ant)\\month1 => x_1=80\times0.94^1=1128\\month2=>x_2=1200\times0.94^2=1060\\month3=>x_3=1200\times0.94^3=996\\month4=>x_4=1200\times0.94^4=936

b.

week Number\ \ \ \ \ \ \ \ \ \ \ \ \ \ Mass(g)\\week1 => x_1=80\times1.1^1=88g\\week2=>x_2=80\times1.1^2=96.8g\\week3=>x_3=80\times1.1^3=106.48g\\week4=>x_4=80\times1.1^4=117.128g

Step-by-step explanation:

a. From the information provided, we can deduce that the population death's follows a Geometric sequence in the form (a,ar,ar^2,ar^3...) where a-first \ term and r-common \ ratio

#Since the population is reducing, r can is obtained as r=1-r=0.94

#The n^t^h term is obtained using the formula x_n=ar^(^n^-^1^), given a=1200

The number of ants alive after every month (in first 4 months)

Month\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Pop(whole \ ant)\\month1 => x_1=80\times0.94^1=1128\\month2=>x_2=1200\times0.94^2=1060\\month3=>x_3=1200\times0.94^3=996\\month4=>x_4=1200\times0.94^4=936

The ant's alive after 4 months is obtained as the value of x_5

x_n=ar^(^n^-^1^)\\1-x_5=1-1200\times 0.94^4=936.89\\\approx 936

Hence, 936 ants are alive after 4 months.

b. As with the above question, the kitten population follows a geometric sequence: (a,ar,ar^2,ar^3...).

#Since it's a growing population , the common ration is the sum of 100% + the growth rate,

r=1.1 and a=80 and x_n=ar^(^n^-^1^)

The population after 4weeks will be:

week Number\ \ \ \ \ \ \ \ \ \ \ \ \ \ Mass(g)\\week1 => x_1=80\times1.1^1=88g\\week2=>x_2=80\times1.1^2=96.8g\\week3=>x_3=80\times1.1^3=106.48g\\week4=>x_4=80\times1.1^4=117.128g

8 0
3 years ago
Two mechanics worked on a car. The first mechanic worked for 15
nevsk [136]

Answer: For the sum of 130

First: $90

Second: $40

Step-by-step explanation:

We write equations for each part of this situation.

<u>The Total Charge</u>

Together they charged 1550. This means 1550 is made up of the first mechanics rate for 15 hours and the second's rate for 5 hours. Lets call the first's rate a, so he charges 15a. The second's let's call b. He charges 5b. We add them together 15a+5b=1550.

<u>The Sum of the Rates</u>

Since the first's rate is a and the second is b, we can write a+b=130 since their sum is 130.

We solve for a and b by substituting one equation into another. Solve for the variable. Then substitute the value into the equation to find the other variable.

For a+b=130, rearrange to b=130-a and substitute into 15a+5b=1550.

15a + 5 (130-a)=1550

15a+650-5a=1550

10a+650-650=1550-650

10a=900

a=$90 was charged by the first mechanic.

We substitute to find the second mechanic's rate.

90+b=130

90-90+b=130-90

b= $40 was charged by the second mechanic

5 0
3 years ago
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