How would I write log(6)27-2log(6)3 as a single logarithm

Question

How would I write log(6)27-2log(6)3 as a single logarithm

Would I start by getting rid of the (-2)- I know its the same log6 but what would I do with the 27 and the 3
I'm really confused on this one and I have to solve several similar ones with no clue on how to it-I've reread my notes but I'm still mixed up
Thanks for any help you can provide

Answer 1

I believe that will be

log₍₆₎(27/3^2). Perhaps Reiny, or Mathmate, or Bob Pursley will double check it. It has been so long since I did one of these.

Answer 2

would 27/3^2 = 3 then in the equation so I would write it as log(6)3

Thanks for trying to help-I appreciate it-maybe I can repost for mathmate or Dr. Bob or Reiny

Answer 3

Please check the above problem-Thank you

Answer 4

Dr. Bob was right.

log(6)27-2log(6)3
=log₆(27) - log₆3²
=log₆(27/3²)
=log₆3

Appendix:
Laws of exponents:
x^a*x^b = x^a+b
x^a/x^b = x^a-b
(x^a)^b = x^ab
x^-a = 1/x^a
x^1/a = ath root of x

Answer 5

To write log(6)27-2log(6)3 as a single logarithm, we can use the properties of logarithms. Specifically, we can use the rule: log(b)x - log(b)y = log(b)(x/y).

First, we'll rewrite the expression using this rule:

log(6)27-2log(6)3 = log(6)(27/3^2)

Next, we simplify the numerator:

27/3^2 = 27/9 = 3

Now, our expression becomes:

log(6)(3)

So, log(6)27-2log(6)3 can be written as log(6)(3).

To solve similar problems, it's important to remember the properties of logarithms. Here are some key rules:

1. log(b)(xy) = log(b)x + log(b)y
2. log(b)(x/y) = log(b)x - log(b)y
3. log(b)(x^n) = n*log(b)x

By applying these rules, you can manipulate the logarithmic expressions to simplify or rewrite them in different forms. It's also helpful to practice solving various logarithmic equations to familiarize yourself with the process.

If you are still confused or struggling with the concept, you might consider seeking additional resources like textbooks, online tutorials, or asking your teacher for clarification.