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 :-

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

$rArrx log_"7"7 =log_"7"15$

$:.x=log_"7"15$

How do you solve 4.7^ { x } - 60= 04.7^ { x } - 60= 0?