How do you solve #3^(4logx) = 5#?

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Answer 1 · 2016-10-29T19:31:07

#x = 10^(log5/log81)#

#log(3^(4logx)) = log5#

#(4logx)log3 = log5#

#4logx = log5/log3#

#logx = (log5/log3)/4#

#logx = log5/log81#

#x = 10^(log5/log81)#

Hopefully this helps!