How do you solve #6(2^(3x-1))-7=9#?

1 Answer
Jim G.
Jun 23, 2017

#x~~0.805" to 3 decimal places"#

Explanation:

#"using the "color(blue)"law of logarithms"#

#• logx^nhArrnlogx#

#rArr6(2^(3x-1))=16larr" add 7 to both sides"#

#rArr2^(3x-1)=16/6=8/3larr" divide both sides by 6"#

#"take ln (natural log of both sides")#

#rArrln2^(3x-1)=ln(8/3)#

#rArr(3x-1)ln2=ln(8/3)larr" use above law"#

#rArr3x=(ln(8/3)/(ln2))+1#

#rArrx=(ln(8/3)/(ln2)+1)/3~~0.805" to 3 dec. places"#

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