Need help solving. I'm stuck at this part, do I change it to log form next?

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Answer 1 · 2017-11-27T06:20:05

#x=7/10#

#(1/5^2)# is the same as saying #(5^-2)# , so you can then write the equation as:

#(5^-2)^(2-5x) = 5^3#

#5^(-4+10x) = 5^3#

#log 5^(-4+10x) = log5^3#

#-4+10x = 3#

#10x = 7#

#x=7/10#