2logan = loga18 + loga ( n - 4 )

Nov 2009
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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)

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Feb 2009
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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