Question

1)Tìm x và y biết:  \(2^x=4^{y-1}\)và \(27^y=3^{x+8}\)

2)Tìm n:

a)\(\left(-\frac{1}{3}\right)^{n-5}=\frac{1}{81}\)   

b)\(2^{-1}.2^n+4.2^n=9.2^5\)  

Answer 1

1. \(2^x=4^{y-1}\Rightarrow2^x=\left(2^2\right)^{y-1}=2^{2y-2}\Rightarrow x=2y-2\)

\(27^y=3^{x+8}\Rightarrow\left(3^3\right)^y=3^{x+8}\Rightarrow3^{3y}=3^{x+8}\Rightarrow3y=x+8\)

ta có: x=2y-2

mà 3y=x+8

=> 3y=2y-2+8

=> 3y-2y+2-8=0

=> y-6=0

=> y=6

x=2y-2

=> x=2.6-2=12-2=10

Vậy x=10; y=6.

2.a.\(\left(-\frac{1}{3}\right)^{n-5}=\frac{1}{81}\)

\(\Rightarrow \left(-\frac{1}{3}\right)^{n-5}=\left(-\frac{1}{3}\right)^4\)

=> n-5=4

=> n=4+5

=> n=9

b.\(2^{-1}.2^n+4.2^n=9.2^5\)

\(\Rightarrow2^n.\left(2^{-1}+4\right)=9.32\)

=> 2ⁿ.(2^-1+4)=288

=> 2ⁿ.(1/2+4)=288

=> 2ⁿ.9/2=288

=> 2ⁿ=288:9/2

=> 2ⁿ=64

=> 2ⁿ=2⁶

Vậy n=6.