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

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The discussion revolves around solving logarithmic equations derived from a math exam problem. The equations in question are y = 2000 x 0.95^x and y = 3000 x 0.90^x. Participants suggest taking logarithms to simplify the equations, specifically focusing on the relationship between the two equations. There is confusion regarding the notation and the steps involved in manipulating logarithms, particularly when an x variable appears in the exponent. The conversation emphasizes the importance of keeping the same base when applying logarithmic properties to solve the equations.
Procrastinate
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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
 
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Hi Procrastinate! :smile:

(try using the X2 tag just above the Reply box :wink:)
Procrastinate said:
y = 2000 x 0.95^x - equation 1
y = 3000 x 0.90^x - equation 2

So 0.95x = 3/2 0.90x …

ok, now take logs. :smile:​
 
tiny-tim said:
So 0.95x = 3/2 0.90x …

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.
 
Procrastinate said:
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.

ah … you need this equation …

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

and what's loga(bc) ? :smile:​
 
tiny-tim said:
ah … you need this equation …

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

and what's loga(bc) ? :smile:​

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

They're two separate questions …

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

what's loga(b2) ?
 
tiny-tim said:
They're two separate questions …

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

what's loga(b2) ?

2logab
 
(just got up :zzz: …)

Yup! :smile:

Now what's loga(bc) ? :wink:
 
clogab
 
  • #10
Yes!

So if you take logs of 0.95x = 3/2 0.90x , you get … ? :smile:
 
  • #11
xlog0.953/2x0.9 = xlog0.92/3x0.95
 
  • #12
Procrastinate said:
xlog0.953/2x0.9 = xlog0.92/3x0.95

I'm not following that at all :confused:

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

take them in the same base (preferably 10 :wink:).
 

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