1/ 3^x-1 + 5.3^x-1 = 162

3^x-1(1 + 5) = 162

3^x-1 = \(\frac{162}{6}\)

3^x-1 = 27

3^x-1 = 3³

x - 1 = 3

x = 4

2/ B = 3¹⁰⁰ - 3⁹⁹ + 3⁹⁸ - 3⁹⁷ + ... + 3² - 3 + 1

\(\Rightarrow\) 3B = 3.3¹⁰⁰ - 3.3⁹⁹ + 3.3⁹⁸ - 3.3⁹⁷ + ... + 3.3² - 3.3 + 3.1

= 3¹⁰¹ - 3¹⁰⁰ + 3⁹⁹ - 3⁹⁸ + ... + 3³ - 3² + 3

Ta có:

4B = 3B + B = (3¹⁰¹ - 3¹⁰⁰ + 3⁹⁹ - 3⁹⁸ + ... + 3³ - 3² + 3) + (3¹⁰⁰ - 3⁹⁹ + 3⁹⁸ - 3⁹⁷ + ... + 3² - 3 + 1)

= 3¹⁰¹ + 3¹⁰⁰ - 3¹⁰⁰ + 3⁹⁹ - 3⁹⁹ + 3⁹⁸ - 3⁹⁸ + ... + 3 - 3 + 1

= 3¹⁰¹ + 1

\(\Rightarrow\) B = \(\frac{3^{101}+1}{4}\)

a) \(2^{x+1}-2^x=32\)

\(\Rightarrow2^x\left(2-1\right)=2^5\)

\(\Rightarrow2^x.1=2^5\)

\(\Rightarrow x=5\)

b) \(2^{x+1}=2\)

\(\Rightarrow2^{x+1}=2^1\)

\(\Rightarrow x+1=1\)

\(\Rightarrow x=0\)

\(2^{x+1}-2^x=32\)

\(2^x.2-2^x=32\)

\(2^x\left(2-1\right)=32\)

\(2^x=32\)

\(x=5\)

\(2^{2x+1}=\left(2^2\right)^x.2=4^x.2\)

1. 2^x=16\(\Rightarrow\)X=4

2. 2^2x-1=2⁷

\(\Rightarrow\)2⁷=2^2.4-1

Vậy x =4

x=4 nha chị

\(2^{x+1}-2^x=32\\ \Rightarrow2^x\cdot2^1-2^x=32\\ \Rightarrow2^x\left(2-1\right)=32\\ \Rightarrow2^x=32:1\\ \Rightarrow2^x=32\\ \Rightarrow2^x=2^5\\ \Rightarrow x=5\)

\(2^{x+1}-2^x=32\\ =>2^x.\left(2^1-1\right)=32\\ =>2^x.1=2^5\\ =>x=5\)

\(2^x+2^{x+2}=32.\left(2^2+1\right)\)

\(\Rightarrow2^x+2^{x+2}=32.5\)

\(\Rightarrow2^x+2^{x+2}=160\)

\(\Rightarrow2^x\left(1+4\right)=160\)

\(\Rightarrow2^x.5=160\)

\(\Rightarrow2^x=160:5=32\)

\(\Rightarrow2^x=2^5\)

\(\Rightarrow x=5\)

Vậy x = 5

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