1. Given that 2^a = 3^b = 18^c, Show that ab=c(b+2a).
2. Given that p^1/2 - p^(-1/2) = 2, Show that p + p^(-1) = 6
Okay, for the next one, I'm not sure if this is the right method, but it's quite long.
$\displaystyle 2^a = 3^b = 18^c$
$\displaystyle 2^a = 3^b = 2^c3^{2c}$
$\displaystyle ln(2^a) = ln(3^b) = ln(2^c3^{2c})$
$\displaystyle ln(2^a) = ln(3^b) = ln(2^c) + ln(3^{2c})$
$\displaystyle a ln2 = bln3 = c ln2 + 2c ln3$
[From here, we know that $\displaystyle \dfrac{ln3}{ln2} = \dfrac{a}{b}$]
Divide by ln(2)
$\displaystyle a = b\dfrac{ln3}{ln2} = c + 2c \dfrac{ln3}{ln2}$
We consider only the centre and right parts now and used what we found a little earlier to remove the natural logs.
$\displaystyle b(\dfrac{a}{b}) = c + 2c (\dfrac{a}{b})$
Can you continue now?