Respuesta :

konrad509
[tex]4^{2x}=7^{x-1}\\ 16^x=7^x\cdot7^{-1}\\ \dfrac{16^x}{7^x}=\dfrac{1}{7}\\ \left(\dfrac{16}{7}\right)^x=\dfrac{1}{7}\\ x=\log_{\tfrac{16}{7}}\dfrac{1}{7}\approx-2.35389[/tex]