Question

How do you solve `2log x = 4log 7`?

Answer 1

`x=49`

Explanation:

Remember:
`color(white)("XXX")a*log(c) = log(c^a)`

Therefore
`color(white)("XXX")2log(x)=4log(7)`

`rarrcolor(white)("XXX")log(x^2)=log(7^4)`

`rarrcolor(white)("XXX")x^2=7^4`

`rarrcolor(white)("XXX")x=7^2=49`