Question

Please help:

Answer 1

$5^{x-1}+5*0.2^{x-2}=26\\ 5^{x-1}+ \dfrac{1}{5^{x-3}} =26\\ 25*5^{x-3}+\dfrac{1}{5^{x-3}} =26\\ Assume\ y=5^{x-3}\\ 25y+ \dfrac{1}{y}=26\\ 25y^2-26y+1=0\\ \Delta=(-26)^2-4*25=576=24^2\\ y= \dfrac{26+24}{50} =1 \ or\ y= \dfrac{26-24}{50} = \dfrac{1}{25} \\ if\ y=1\ then 5^{x-3}=1==\ \textgreater \ x-3=0==\ \textgreater \ x=3\\ if\ y= \dfrac{1}{25}\ then\ 5^{x-3}=\dfrac{1}{5^2}==\ \textgreater \ x-1=0==\ \textgreater \ x=1$