# Log/Index question which I am stumped with.

1. Nov 20, 2009

### Procrastinate

I had this question in the problem solving part of my Maths Exam. When I extrapolated all the critical information from the question, I narrowed it down to this equation. However, I just couldn't solve the equation. Could anyone please help? This has been bugging me for the whole day.

y = 2000 x 0.95^x - equation 1
y = 3000 x 0.90^x - equation 2

2. Nov 20, 2009

### tiny-tim

Hi Procrastinate!

(try using the X2 tag just above the Reply box )
So 0.95x = 3/2 0.90x

ok, now take logs.

3. Nov 20, 2009

### Procrastinate

Wouldn't that be log 0.95 3/20.9x = x

That's what I did, but the fact that there was an x still in the index was what hindered me from completing it.

4. Nov 20, 2009

### tiny-tim

ah … you need this equation …

keeping to the same base, a, what's loga(b2) ?

and what's loga(bc) ?

5. Nov 20, 2009

### Procrastinate

Sorry, but I am still a bit confused about the notation. Do I square it first and then do loga(bc)?

6. Nov 20, 2009

### tiny-tim

They're two separate questions …

I thought I'd start you off with an easy one first …

what's loga(b2) ?

7. Nov 20, 2009

### Procrastinate

2logab

8. Nov 21, 2009

### tiny-tim

(just got up :zzz: …)

Yup!

Now what's loga(bc) ?

9. Nov 21, 2009

### Procrastinate

clogab

10. Nov 21, 2009

### tiny-tim

Yes!

So if you take logs of 0.95x = 3/2 0.90x , you get … ?

11. Nov 21, 2009

### Procrastinate

xlog0.953/2x0.9 = xlog0.92/3x0.95

12. Nov 21, 2009

### tiny-tim

I'm not following that at all

This is supposed to be taking logs of 0.95x = 3/2 0.90x

take them in the same base (preferably 10 ).