How do you solve #5^(4x) = 2115#?

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Answer 1 · 2016-01-04T17:38:17

Use #loga^b=b*loga#

#x=1.18936#

#5^(4x)=2115#

#log5^(4x)=log2115#

#4x*log5=log2115#

#x=(log2115)/(4*log5)#

#x=1.18936#