Solve the equation (linear equation)[tex]8^{2x+7} = (\frac{1}{32})^{3x}[/tex]

Answer: [tex]x=-1[/tex]Step-by-step explanation: By the negative exponent rule, you have that: [tex](\frac{1}{a})^n=a^{-n}[/tex] By the exponents properties, you know that: [tex](m^n)^l=m^{(nl)}[/tex] Therefore, you can rewrite the left side of the equation has following:[tex](\frac{1}{8})^{-(2x+7)}=(\frac{1}{32})^{3x}[/tex]  Descompose 32 and 8 into its prime factors: [tex]32=2*2*2*2*2=2^5\\8=2*2*2=2^3[/tex] Rewrite: [tex](\frac{1}{2^3})^{-(2x+7)}=(\frac{1}{2^5})^{3x}[/tex] Then: [tex](\frac{1}{2})^{-3(2x+7)}=(\frac{1}{2})^{5(3x)}[/tex] As the base are equal, then: [tex]-3(2x+7)=5(3x)[/tex] Solve for x: [tex]-6x-21=15x\\-21=15x+6x\\-21=21x\\x=-1[/tex]