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vivado [14]
3 years ago
12

Solve (1/81)^x = 243^x-9

Mathematics
1 answer:
user100 [1]3 years ago
4 0

Answer:

x=5

Step-by-step explanation:

(1/81)^x=(243)^x-9

(1/3^4)=3^5(x-9)

3^-4x=3^5(x-9)

-4x=5(x-9)

-4x=5x-45

-4x-5x=-45

-9x=-45

x=5

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NEED ANSWERED ASAP WILL REWARD BRAINLIEST
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Answer:

The sum of the first 100 terms is 60400

Step-by-step explanation:

* Lets revise the arithmetic sequence

- There is a constant difference between each two consecutive

  numbers

- Ex:

# 2  ,  5  ,  8  ,  11  ,  ……………………….

# 5  ,  10  ,  15  ,  20  ,  …………………………

# 12  ,  10  ,  8  ,  6  ,  ……………………………

* General term (nth term) of an Arithmetic sequence:

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- Un = a + (n – 1)d, where a is the first term , d is the difference

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 number

- The sum of first n terms of an Arithmetic sequence is calculate from

 Sn = n/2[a + l], where a is the first term and l is the last term

* Now lets solve the problem

- We will use method (1)

- From the table the terms of the sequence are:

 10 , 22 , 34 , 46 , 58 , 82 , 94 , ............., where 10 is the first term

∵ an = a1 + (n - 1) d ⇒ explicit formula

∵ a1 = 10 and a2 = 22

∵ d = a2 - a1

∴ d = 22 - 10 = 12

- The 100th term means the term of n = 100

∴ a100 = 10 + (100 - 1) 12

∴ a100 = 10 + 99 × 12 = 10 + 1188 = 1198

∴ The 100th term is 1198

- Lets find the sum of the first 100 terms of the sequence

∵ Sn = n/2[a1 + an]

∵ n = 100 , a = 10 , a100 = 1198

∴ S100 = 100/2[10 + 1198] = 50[1208] = 60400

* The sum of the first 100 terms is 60400

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2 years ago
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