Question

1. \(\left( x + \frac{1}{2} \right)^2 = \frac{1}{9}\)

2. \(\left( 3x - \frac{1}{4} \right)^3 = -\frac{1}{64}\)

3. \(\left( x - \frac{1}{3} \right)^2 + \left( x^2 - \frac{1}{9} \right)^4 = 0\)

4. \(2^{x+2} - 2^x = 96\)

6. \(7^{x+2} + 2 \cdot 7^{x-1} = 345\)

7. \(\left( \frac{1}{3} \right)^{2x-1} = 243\)

8. \( (0.125)^{n-1} = 64\)

Answer 1

1: \(\left(x+\frac12\right)^2=\frac19\)

=>\(\left[\begin{array}{l}x+\frac12=\frac13\\ x+\frac12=-\frac13\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac13-\frac12=-\frac16\\ x=-\frac13-\frac12=-\frac56\end{array}\right.\)

2: \(\left(3x-\frac14\right)^3=-\frac{1}{64}\)

=>\(\left(3x-\frac14\right)^3=\left(-\frac14\right)^3\)

=>\(3x-\frac14=-\frac14\)

=>3x=0

=>x=0

3: \(\left(x-\frac13\right)^2+\left(x^2-\frac19\right)^4=0\)

=>\(\begin{cases}x-\frac13=0\\ x^2-\frac19=0\end{cases}\Rightarrow\begin{cases}x=\frac13\\ x^2=\frac19\end{cases}\)

=>\(x=\frac13\)

4: \(2^{x+2}-2^{x}=96\)

=>\(2^{x}\cdot4-2^{x}=96\)

=>\(2^{x}\cdot3=96\)

=>\(2^{x}=\frac{96}{3}=32=2^5\)

=>x=5

5: \(7^{x+2}+2\cdot7^{x-1}=345\)

=>\(7^{x-1}\cdot7^3+2\cdot7^{x-1}=345\)

=>\(7^{x-1}\left(7^3+2\right)=345\)

=>\(7^{x-1}\left(343+2\right)=345\)

=>\(7^{x-1}=1\)

=>x-1=0

=>x=1

7: \(\left(\frac13\right)^{2x-1}=243\)

=>\(\left(\frac13\right)^{2x-1}=\left(\frac13\right)^{-5}\)

=>2x-1=-5

=>2x=-4

=>x=-2

8: \(0,125^{n-1}=64\)

=>\(\left(\frac18\right)^{n-1}=\left(\frac18\right)^{-2}\)

=>n-1=-2

=>n=-2+1=-1