Question

How do you solve `log_4 7 + 2 log_4 x = log_4 2`?

Answer 1

`x=sqrt(2/7)`

Explanation:

Given:
`color(white)("XXX")log_4(7)+2log_4(x)=log_4(2)`

Remember
`color(white)("XXX")`log multiplication rule: `log_b(p*q) =log_b(p)+log_b(q)`
`color(white)("XXX")`log power rule: `log_b(s^t) = t*log_b(s)`

Therefore the given equation can be rewritten as
`color(white)("XXX")log_4(7x^2)=log_4(2)`

from which it follows that
`color(white)("XXX")7x^2=2`

`color(white)("XXX")x^2=2/7`

`color(white)("XXX")x=sqrt(2/7)`
`color(white)("XXXXXX")`we can ignore the negative root as extraneous since `sqrt(x)` requires `x>=0` for Real solutions