How Do I Solve These Logarithmic Equations From My Math Exam?

[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

 

[tiny-tim]

Hi Procrastinate!

(try using the X² tag just above the Reply box )

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

So 0.95^x = 3/2 0.90^x …

ok, now take logs. ​

 

[Procrastinate]

So 0.95^x = 3/2 0.90^x …

Wouldn't that be log _0.95 3/20.9^x = 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.

 

[tiny-tim]

Wouldn't that be log _0.95 3/20.9^x = 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.

ah … you need this equation …

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

and what's log_a(b^c) ? ​

 

[Procrastinate]

ah … you need this equation …

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

and what's log_a(b^c) ? ​

Sorry, but I am still a bit confused about the notation. Do I square it first and then do log_a(b^c)?

 

[tiny-tim]

Sorry, but I am still a bit confused about the notation. Do I square it first and then do log_a(b^c)?

They're two separate questions …

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

what's log_a(b²) ?

 

[Procrastinate]

They're two separate questions …

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

what's log_a(b²) ?

2log_ab

 

[tiny-tim]

(just got up :zzz: …)

Yup!

Now what's log_a(b^c) ?

 

[Procrastinate]

clog_ab

 

[tiny-tim]

Yes!

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

 

[Procrastinate]

xlog_0.953/2x0.9 = xlog_0.92/3x0.95

 

[tiny-tim]

xlog_0.953/2x0.9 = xlog_0.92/3x0.95

I'm not following that at all

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

take them in the same base (preferably 10 ).