1. ## Exponents

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

2. Could you tell us what you came up with so far?

But for the first part, I don't understand where the c comes from...

3. Yeah sure....

For Question 2,

p^1/2 - p^(-1/2) = 2

√p - 1/√p = 2

Let y = √p

y - 1/y = 2

y^2 - 2y - 1 = 0

I can do this question up to here....Then,I'm blurr......

4. {p^1/2 - p^(-1/2)}^2=2^2

5. For the second one, square both sides.

For the first one, are you sure this is the right question?

6. Sorry Unknown008

7. The 1st question is the right one.....I missed the 'c' earlier....sorry!

8. 18^c=(2*3^2)^c=2^c*3^2c

9. For the second one,

I'm not sure what to do after squaring it.

10. just "play" with the equations... and try to enjoy by doing it...

11. You got:

$(\sqrt{p} - \dfrac{1}{\sqrt{p}})^2 = 2^2$

This becomes:

$\sqrt{p}^2 - 2(\dfrac{1}{\sqrt{p}})(\sqrt{p}) + (\dfrac{1}{\sqrt{p}})^2 = 4$

$p - 2 + \dfrac{1}{p} = 4$

12. Okay,I managed to get until p^2 - 6p + 1 = 0 ......Then how to show that p + p^(-1) = 6

13. From where I left, just move the '-2' and what do you get?

14. Okay got it.....Thanks!!

15. Okay, for the next one, I'm not sure if this is the right method, but it's quite long.

$2^a = 3^b = 18^c$

$2^a = 3^b = 2^c3^{2c}$

$ln(2^a) = ln(3^b) = ln(2^c3^{2c})$

$ln(2^a) = ln(3^b) = ln(2^c) + ln(3^{2c})$

$a ln2 = bln3 = c ln2 + 2c ln3$

[From here, we know that $\dfrac{ln3}{ln2} = \dfrac{a}{b}$]

Divide by ln(2)

$a = b\dfrac{ln3}{ln2} = c + 2c \dfrac{ln3}{ln2}$

We consider only the centre and right parts now and used what we found a little earlier to remove the natural logs.

$b(\dfrac{a}{b}) = c + 2c (\dfrac{a}{b})$

Can you continue now?

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