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

1. ## 2logan = loga18 + loga ( n - 4 )

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)

2. Originally Posted by 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)
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

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