How do you solve #4.7^ { x } - 60= 0#?

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Answer 1 · 2018-04-16T15:02:27

#log_"7"15#

#4.7^x-60=0#

#rArr4.7^x=60#

#rArr7^x=15#

Taking logarithm to the base #7# on both sides we get :-

#rArrlog_"7"7^x=log_"7"15#

#rArrx log_"7"7 =log_"7"15#

#:.x=log_"7"15#