Logs question: Given that $\log_a(x) = 2(\log_a(k)-\log_a(2))$, showing that $k^2-4x=0$

Question

Logs question: Given that $\log_a(x) = 2(\log_a(k)-\log_a(2))$, showing that $k^2-4x=0$.

Thanks in advance.

EDIT: Managed to solve it.

$\ log_a(x) = 2log_a(k)-2log_a(2)$
$\ log_a(x) = log_a(k^2)-log_a(4)$
$\ log_a(x) = log_a(k^2 / 4)$
$\ x = k^2/4$
$\ 4x = k^2$
$\ k^2 -4x =0$

Didn't see all the replies, thanks anyway :)

Answer 1

HINT:

Raise $a$ to the left-hand side and to the right-hand side and set them as equal. Use properties of powers and of logarithms to arrive at your desired conclusion!

It may be worth remembering that $a^{\log_a{Y}} = Y$

Answer 2

$\log_a(x) = 2(log_a(k)-log_a(2))$

$\log_a(x) = 2(log_a(\frac{k}{2}))$

$\log_a(x) = log_a(\frac{k}{2})^2$

$x = (\frac{k}{2})^2$

$x = \frac{k^2}{4}$

$4x = k^2$

$k^2 - 4x = 0$

Some Formulas

1. $log_a(x) - log_a(y) = log_a(\frac{x}{y})$

2. $log_a(x) + log_a(y) = log_a(xy)$

3. $ylog_a(x) = log_a(x)^y$

Answer 3

Hint: Use $\log x + \log y = \log xy$ and $c \log x = \log (x^c)$.

Answer 4

we have after the rules of logarithm $$2(\log_a k-\log_a 2)=2\log_a\left(\frac{k}{a}\right)$$ and if $$k>0$$ $$\log_a x=\log_a \left(\frac{k}{2}\right)^2$$ thus we get $$x=\left(\frac{k}{2}\right)^2$$