Question

Solve the equation (linear equation) %20%3D%28%5Cfrac%7B3%7D%7B4%7D%29%5E%7Bx-7%7D" id="TexFormula1" title="(\frac{16}{9}) ^{2x +5} =(\frac{3}{4})^{x-7}" alt="(\frac{16}{9}) ^{2x +5} =(\frac{3}{4})^{x-7}" align="absmiddle" class="latex-formula">

Answer 1

Answer: $x=-\frac{3}{5}$

Step-by-step explanation:

By the negative exponent rule, you have that:

$(\frac{1}{a})^n=a^{-n}$

By the exponents properties, you know that:

$(m^n)^l=m^{(nl)}$

You can rewrite 16 and 9 as following:

16=4²

9=3²

Therefore, you can rewrite the left side of the equation has following:

$(\frac{4^2}{3^2})^{(2x+5)}=(\frac{3}{4})^{(x-7)}\\\\(\frac{3^2}{4^2})^{-(2x+5)}=(\frac{3}{4})^{(x-7)}\\\\(\frac{3}{4})^{-2(2x+5)}=(\frac{3}{4})^{(x-7)}$

 As the base are equal, then:

$-2(2x+5)=x-7$

Solve for x:

$-2(2x+5)=x-7\\-4x-10=x-7\\-4x-x=-7+10\\-5x=3\\x=-\frac{3}{5}$