How do you solve #6(2^(3x-1))-7=9#?
1 Answer
Jun 23, 2017
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"#
