Question

How do you solve `log_5(log3x) = 0`?

Answer 1

Use the equivalence:
`color(white)("XXX")color(red)(log_b(a)=c) hArr color(blue)(b^c=a)`
to determine
`color(white)("XXX")x=10/3`

Explanation:

`log_5(log (3x))=0`

`rArr 5^0 = log(3x)`
`rarrcolor(white)("XXX")log(3x)=1`

`log(3x)=1`
`rArr 10^1=3xcolor(white)("XXXXXXXXXXXXX")`remember the default `log` base is `10`
`rarrcolor(white)("XXX")x=10/3`