How do you solve #8(10^(3x))=12#?

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Answer 1 · 2017-02-04T15:20:25

#x=0.0587#.

We will use the familiar rules of Log function.

#8(10^(3x))=12#

#rArr 10^(3x)=12/8=3/2#.

#rArr log_10 10^(3x)=log_10 (3/2)=log_10 3- log_10 2#.

#rArr (3x)log_10 10=0.4771-0.3010=0.1761.#

#rArr 3x=0.1761#.

#rArr x=0.1761/3#.

#:." The Soln. is, "x=0.0587#.