How do you solve #3(5^(x-1))=21#?

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Answer 1 · 2018-07-27T10:20:38

#x=\frac{\ln35}{\ln5}#

Given that

#3(5^{x-1})=21#

#5^{x-1}=21/3#

#5^{x-1}=7#

Taking logs on both the sides as follows

#\ln5^{x-1}=\ln 7#

#(x-1)\ln 5=\ln7#

#x-1=\frac{\ln 7}{\ln 5}#

#x=\frac{\ln 7}{\ln 5}+1#

#x=\frac{\ln 7+\ln 5}{\ln 5}#

#x=\frac{\ln35}{\ln5}#

Answer 2 · 2018-07-27T10:21:41

#x=2.21# (2.d.p)

#3(5^(x-1))=21#

#5^(x-1)=7#

#log_10 5^(x-1)=log_10 7#

#(x-1) log_10 5=log_10 7#

#x-1=log_10 7/log_10 5#

#x=1+log_10 7/log_10 5#

#x=2.21# (2.d.p)