Question

Solve the following equations.
${32}^{\frac{x+5}{x-7}}=0.25\cdot {128}^{\frac{x+17}{x-3}}$

Answer 1

${32}^{\frac{x+5}{x-7}}={\mathrm{0.25.128}}^{\frac{x+17}{x-3}}$

$(2^5{)}^{\frac{x+5}{x-7}}=1/4.(2^7{)}^{\frac{x+17}{x-3}}$

$2^2{.2}^{5\left(\frac{x+5}{x-7}\right)}=2^{7.\left(\frac{x+17}{x-3}\right)}$

$2^{2+5.\left(\frac{x+5}{x-7}\right)}=2^{7\left(\frac{x+17}{x-3}\right)}$

$2+5\left(\frac{x+5}{x-7}\right)=7\left(\frac{x+17}{x-3}\right)$

$2+\frac{5x+25}{x-7}=\frac{7x+119}{x-3}$

$\frac{2x-14+5x+25}{x-7}=\frac{7x+119}{x-3}$

$\frac{7x+11}{x-7}=\frac{7x+119}{x-3}$

$(x-3)(7x+11)=(x-7)(74+119)$

$7x^2+11x-21x-33=7x^2+119x-49x-833$

$-10x-33=70x-833$

$800=80x$

$x=10$