Question

Tìm x ϵ N, biết:

a) 2^x + 2^x+1 = 96

b) 3^4x+4 = 81^x+3

Answer 1

\(a,2^x+2^{x+1}=96\)

\(\Rightarrow2^x+2^x.2=96\)  \(\Rightarrow2^x\left(1+2\right)=96\)

\(\Rightarrow2^x.3=96\)  \(\Rightarrow2^x=32\)

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

\(b,3^{4x+4}=81^{x+3}\)

\(\Rightarrow3^{4x+4}=3^{4x+12}\)

\(\Rightarrow4x+4=4x+12\)  (Vô lý)

Vậy \(x\in\varnothing\)

 

Answer 2

a/ \(2^x+2^{x+1}=96\)

\(2^x+2^x.2=96\)

\(2^x\cdot\left(2+1\right)=96\)

\(2^x=\frac{96}{3}=32\)

\(2^x=2^5\)

\(=>x=5\)

b/ \(3^{4x+4}=81^{x+3}\)

\(\Rightarrow3^{4x+4}-81^{x+3}=0\)

\(3^{4x}.3^4-3^{4x}\cdot81^3=0\)

\(3^{4x}\cdot\left(81-81^3\right)=0\)

\(3^{4x}=\frac{0}{81-81^3}\)

\(3^{4x}=0\Rightarrow x=0\)

Answer 3

a) 2^x + 2^x x 2¹= 96

2^x x (1+ 2)=96

2^x = 96 : 3= 32 =2⁵

x= 5