# 2logan = loga18 + loga ( n - 4 )

#### ansonbound

Given that:

2logan = loga18 + loga ( n - 4 )

find the possible values of n.

>>how do i get to logan^2 = loga 18(n-4)

is it correct that

2logan = 2 logan^2

how do you get from:

loga18 + loga ( n - 4 )

to logan^2 = loga 18(n-4)
(Nerd)(Nerd)

#### e^(i*pi)

MHF Hall of Honor
Given that:

2logan = loga18 + loga ( n - 4 )

find the possible values of n.

>>how do i get to logan^2 = loga 18(n-4)

is it correct that

2logan = 2 logan^2

how do you get from:

loga18 + loga ( n - 4 )

to logan^2 = loga 18(n-4)
(Nerd)(Nerd)
I assume you mean base a.

$$\displaystyle 2\log_a(n) = \log_a(n^2)$$ is true as it's the power law.

You can simplify the RHS using the addition law:

$$\displaystyle \log_a(18) + \log_a(n-4) = \log_a[18(n-4)]$$

$$\displaystyle \log_a(n^2) = \log_a[18(n-4)]$$

Now if the two bases are the same then the exponents must be equal.

$$\displaystyle n^2 = 18(n-4)$$

Which is a standard quadratic equation but remember that $$\displaystyle n>0$$ to satisfy the original domain