Question

log7 + log (n - 2) = log 6n

Answer 1

If 7 is the base, log(7) = 1

But the calculation is the same, whatever the base is...

log(7) = log(6n) - log(n-2) = log[6n / (n-2)]

As the application log is a bijection :

7 = 6n / (n-2)
7n - 14 = 6n
n = 14

Check : log 7 + log 12 = log 84 OK as 7 x 12 = 84

Hope it helped...

Answer 2

http://www.chilimath.com/algebra/advanced/log/images/rules%20of%20exponents.gif

I'm assuming you're using base10, thus

look at rule 1 on that picture
after that you'd use rule 2

Answer 3

\(\bf log(7) + log (n - 2) = log(6n)\\
\color{blue}{\text{using rule 1 on that picture}}\\
log(7) + log (n-2) \implies log(7(n-2))\\
log(7(n-2))= log(6n)\\
\color{blue}{\text{substract log(6n) from both sides}}\\
\color{blue}{\text{then use rule 2 on that picture}}
\)

Answer 4

I don't understand this at all.