Solve the equation (linear equation)[tex]4^{x-7}[/tex] × [tex]8^{2x-3} =\frac{32}{2^{x-9} }[/tex]

Answer: [tex]x=\frac{37}{9}[/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] [tex](m^n)(m^l)=m^{(n+l)}[/tex] Rewrite 4, 8 and 32 as following: 4=2² 8=2³ 32=2⁵ Rewrite the expression: [tex](2^2)^{(x-7)}*(2^3)^{(2x-3)}=\frac{32}{2^{(x-9)}}[/tex] Keeping on mind the exponents properties, you have: [tex](2)^{2(x-7)}*(2)^{3(2x-3)}=32(2^{-(x-9)}[/tex] [tex](2)^{2(x-7)}*(2)^{3(2x-3)}=(2^5)(2^{-(x-9)})\\\\(2)^{(2x-14)}*(2)^{(6x-9)}=(2^5)(2^{(-x+9)})\\\\2^{((2x-14)+(6x-9))}=2^{(5+(-x+9))}[/tex] As the bases are equal, then: [tex](2x-14)+(6x-9)=5+(-x+9)\\\\2x-14+6x-9=5-x+9\\\\8x-23=14-x\\9x=37[/tex] [tex]x=\frac{37}{9}[/tex]