Question

1) \(4^{x+1}=64^x\)

2) \(\dfrac{x}{2}-\dfrac{x}{5}=1+x\)

Answer 1

\(1,\Rightarrow4^{x+1}=4^{3x}\\ \Rightarrow x+1=3x\\ \Rightarrow2x=1\\ \Rightarrow x=\dfrac{1}{2}\\ 2,\Rightarrow5x-2x=10+10x\\ \Rightarrow7x=-10\\ \Rightarrow x=-\dfrac{10}{7}\)

Answer 2

1. 4^{x + 1} = 64^x

<=> 4^{x + 1} = 4^3x

<=> x + 1 = 3x

<=> 1 = 2x

<=> \(x=\dfrac{1}{2}\)

2. \(\dfrac{x}{2}-\dfrac{x}{5}=1+x\)

<=> \(\dfrac{5x}{10}-\dfrac{2x}{10}=\dfrac{10}{10}+\dfrac{10x}{10}\)

<=> 5x - 2x = 10 + 10x

<=> 5x - 2x - 10x = 10

<=> -7x = 10

<=> \(x=\dfrac{-10}{7}\)