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)
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