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VARVARA [1.3K]
3 years ago
11

Solve the following logarithmic equations. ln(x^6) = 36

Mathematics
1 answer:
Hoochie [10]3 years ago
8 0

Answer:

The solution is x = e⁶

Step-by-step explanation:

Hi there!

First, let´s write the equation

ln(x⁶) = 36

Apply logarithm property: ln(xᵃ) = a ln(x)

6 ln(x) = 36

Divide both sides of the equation by 6

ln(x) = 6

Apply e to both sides

e^(ln(x)) = e⁶

x = e⁶

The solution is x = e⁶

Let´s prove why e^(ln(x)) = x

Let´s consider this function:

y = e^(ln(x))

Apply ln to both sides of the equation

ln(y) = ln(e^(ln(x)))

Apply logarithm property: ln(xᵃ) = a ln(x)

ln(y) = ln(x) · ln(e)         (ln(e) = 1)

ln(y) = ln(x)

Apply logarithm equality rule: if ln(a) = ln(b) then, a = b

y = x

Since y = e^(ln(x)), then x =e^(ln(x))

Have a nice day!

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What is the answer to p+12 =-18
MariettaO [177]

Answer:

p=-30

Step-by-step explanation:

p+12=-18

subtract 12 on both sides and you get p=-30

Hope this helps!!

3 0
3 years ago
Read 2 more answers
In a survey of 400 randomly selected​ people, it was determined that 22 play soccer. What is the probability that a randomly sel
insens350 [35]

Answer:

5.5% probability that a randomly selected person plays​ soccer

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

Desired outcomes:

People surveyed who play soccer, so D = 22

Total outcomes:

Total people surveyed, so P = 400

What is the probability that a randomly selected person plays​ soccer?

P = \frac{D}{T} = \frac{22}{400} = 0.055

5.5% probability that a randomly selected person plays​ soccer

6 0
2 years ago
What is question 32 and can you please explain how to do it.
lukranit [14]

Answer is B.

5√3(5+2√6)=

25√3+10√18=

25√3+30√2

Note: I multiplied 10 by 3, because the prime factors of 18 are 3, 3 and 2 so I made a pair from the 3 and multipled and was left with 2.


5 0
2 years ago
HURRY!! NO SPAM!!!!!!!!!!!!!!!!!!
Zielflug [23.3K]
  • 2moles of NaCl produced by 2 moles of NaClO_3
  • Let required moles be x .

\\ \bull\sf\longmapsto 2:2=3:x

\\ \bull\sf\longmapsto \dfrac{2}{2}=\dfrac{3}{x}

\\ \bull\sf\longmapsto \dfrac{3}{x}=1

\\ \bull\sf\longmapsto x=3mol

8 0
2 years ago
Read 2 more answers
The operations manager of a large production plant would like to estimate the mean amount of time a worker takes to assemble a n
xz_007 [3.2K]

Answer:

(15,595, 16,805)

Step-by-step explanation:

We have to:

m = 16.2, sd = 3.75, n = 150

m is the mean, sd is the standard deviation and n is the sample size.

the degree of freedom would be:

n - 1 = 150 - 1 = 149

df = 149

at 95% confidence level the t is:

alpha = 1 - 95% = 1 - 0.95 = 0.05

alpha / 2 = 0.05 / 2 = 0.025

now well for t alpha / 2 (0.025) and df (149) = t = 1,976

the margin of error = E = t * sd / (n ^ (1/2))

replacing:

E = 1,976 * 3.75 / (150 ^ (1/2))

E = 0.605

The 95% confidence interval estimate of the popilation mean is:

m - E <u <m + E

16.2 - 0.605 <u <16.2 + 0.605

15,595 <u <16,805

(15,595, 16,805)

5 0
3 years ago
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