Question

[𝟸𝟿_𝟸𝟷]

Cần giúp , gấp !

Answer 1

[Xyz OLM]

g) 2.3^x = 162

=> 3^x = 81 

=> 3^x = 3⁴

=> x = 4 

Vậy x=  4

h) (2x - 15)⁵ = (2x - 15)³

<=> (2x - 15)⁵ - (2x - 15)³ = 0

<=> (2x - 15)³[(2x - 15)² - 1] = 0

<=> \(\orbr{\begin{cases}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x-15=0\\\left(2x-15\right)^2=1^2\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=15\\2x-15\in\left\{1;-1\right\}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=7,5\\x\in\left\{8;7\right\}\end{cases}}\)

Vậy \(x\in\left\{7;7,5;8\right\}\)

k) 4.2^x - 3 = 1

=> 4.2^x = 4

=> 2^x = 1

=> x = 0

l) 3^{x + 2} - 5.3^x = 36

<=> 3^x.9 - 5.3^x = 36

=> 3^x.4 = 36

=> 3^x = 9

=> x = 2

m) 7.4^{x - 1} + 4^{x + 1} = 23

<=> 7.4^{x - 1} + 4^{x - 1}.4² = 23

<=> 4^{x - 1}(7 + 4²) = 23

<=> 4^{x - 1} = 1

<=> 4^{x - 1} = 4⁰

<=> x - 1 = 0

<=> x = 1

n) 2.2^2x + 4³.4^x = 1056

<=> 2.2^2x + 64.2^2x = 1056

<=> 66.2^2x = 1056

<=> 2^2x = 16

<=> 2^2x = 2⁴

<=> 2x = 4

<=> x = 2