Question

si 8(2^x) = 5^y+8, entonces y cuando Y=-8, se tiene que x es:

Answer 1

Despejamos:

$8\left(2^x\right)=5^{-8}+8$

Resolvemos:

$8\left(2^x\right)=5^{-8}+8\\ \\ \frac{8\cdot \:2^x}{8}=\frac{5^{-8}}{8}+\frac{8}{8}\\ \\ 2^x=\frac{5^{-8}+8}{8}\\ \\ 2^x=\frac{3125001}{3125000}\\ \\ \ln \left(2^x\right)=\ln \left(\frac{3125001}{3125000}\right)\\ \\ x\ln \left(2\right)=\ln \left(\frac{3125001}{3125000}\right)\\ \\ \boxed{\boldsymbol{x=\frac{\ln \left(\frac{3125001}{3125000}\right)}{\ln \left(2\right)}}}$

Conclusión:

$\boxed{\boldsymbol{x=\frac{\ln \left(\frac{3125001}{3125000}\right)}{\ln \left(2\right)}}}$

¡SUERTE!