Question

Tìm x biết

\(3^x+3^{x+2}=2430\)\(2^{x+3}-2^x=224\)​

Áp dụng công thức \(\left(x.y\right)^n=x^n.y^n\) 

                                \(\left(\frac{x}{y}\right)^n=\frac{x^n}{y^n}\left(y\ne0\right)\)

Answer 1

1. \(3^x+3^{x+2}=2430\)

    \(3^x\left(1+3^2\right)=2430\)

    \(3^x.10=2430\)

    \(3^x=243\)

    \(3^x=3^5\)

    \(x=5\)

2. \(2^{x+3}-2^x=224\)

    \(2^x\left(2^3-1\right)=224\)

    \(2^x.7=224\)

    \(2^x=32\)

    \(2^x=2^5\)

    \(x=5\)

Answer 2

1. 3^x + 3^x+2 = 2430

3^x.1+3^x.3^2=2430

3^x.1+3^x.9=2430

3^x.(1+9)=2430

3^x.10=2430

3^x=2430:10

3^x=243

3^x=3^5

=> x=5

Vậy x =5

2. 2^x+3  - 2^x =224

2^x.2^3-2^x.1=224

2^x.8-2^x.1=224

2^x.(8-1)=224

2^x.7=224

2^x=224:7

2^x=32

2^x=2^5

=> x=5

Vậy x=5