Question

Tìm X :

\(\frac{1}{2}.2^x+2^{x+2}=2^8+2^5\)

\(9^x:3^x=3^7\)

\(7^{x+2}+2.7^{x-1}=345\)

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Answer 1

\(\frac{1}{2}\cdot2^x+2^x\cdot2^2=2^8+2^5\)

\(2^x\left(\frac{1}{2}+4\right)=2^8+2^5\)

\(2^x\cdot\frac{9}{2}=288\)

\(2^x=64\)

\(2^x=2^6\)

\(x=6\)

\(9^x:3^x=3^7\)

\(3^{2x}:3^x=3^7\)

\(3^x=3^7\)

\(x=7\)

\(7^{x+2}+2\cdot7^{x-1}=345\)

\(7^x\cdot7^2+2\cdot7^x:7=345\)

\(7^x\left(7^2+\frac{2}{7}\right)=345\)

\(7^x\cdot\frac{345}{7}=345\)

\(7^x=7\)

\(x=1\)

Answer 2

a) 1/2.2^x + 2^x+2 = 256 + 32

1/2.2^x + 2^2.2^x=288

2^x(1/2+4)= 288

2^x.4,5=288

2^x= 288:4,5

2^x=64=2^6

x=6