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

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Homework Help Overview

The discussion revolves around solving logarithmic equations derived from a math exam problem involving two exponential equations. The original poster presents equations that involve exponential decay factors and seeks assistance in solving for the variable x.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss taking logarithms of the equations to isolate x. There are attempts to manipulate the equations and questions about the notation and properties of logarithms. Some participants express confusion about handling the variable x in the exponent.

Discussion Status

The conversation is ongoing, with participants providing hints and asking questions to clarify logarithmic properties. There is a focus on understanding the manipulation of logarithmic expressions, but no consensus or resolution has been reached yet.

Contextual Notes

Participants are navigating through the complexities of logarithmic notation and properties, with some expressing confusion about specific steps in the process. The discussion reflects a learning environment where assumptions and definitions are being questioned.

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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