Question

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

Answer 1

`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"`