[tex]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\\
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