Answer: [tex]x=-\frac{3}{5}[/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]
You can rewrite 16 and 9 as following:
16=4²
9=3²
Therefore, you can rewrite the left side of the equation has following:
[tex](\frac{4^2}{3^2})^{(2x+5)}=(\frac{3}{4})^{(x-7)}\\\\(\frac{3^2}{4^2})^{-(2x+5)}=(\frac{3}{4})^{(x-7)}\\\\(\frac{3}{4})^{-2(2x+5)}=(\frac{3}{4})^{(x-7)}[/tex]
As the base are equal, then:
[tex]-2(2x+5)=x-7[/tex]
Solve for x:
[tex]-2(2x+5)=x-7\\-4x-10=x-7\\-4x-x=-7+10\\-5x=3\\x=-\frac{3}{5}[/tex]
