Solving 2^x=7*5^x: Tips & Tricks

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Homework Help Overview

The discussion revolves around solving the equation 2^x = 7 * 5^x, which involves logarithmic properties and manipulation. Participants are exploring the application of logarithms to both sides of the equation.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss taking the natural logarithm of both sides and express confusion about applying logarithmic properties, particularly with the right side of the equation. There are attempts to clarify the use of logarithmic identities.

Discussion Status

Some participants have provided guidance on using logarithmic properties, such as ln(ab) = ln(a) + ln(b). There appears to be a mix of understanding and confusion, with some participants still questioning their approach.

Contextual Notes

One participant mentions struggling with the manipulation of logarithms, indicating a potential gap in understanding logarithmic properties. There is also a reference to a hint that may not be fully utilized, suggesting constraints in the discussion.

chjopl
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2^x=7*5^x
I know you have to take ln of both sides but i am having trouble doing that for the right side.
 
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Remember the property of logarithms, ln(ab)=ln(a)+ln(b). That should help you deal with the right hand side.
 
Last edited:
chjopl said:
2^x=7*5^x
I know you have to take ln of both sides but i am having trouble doing that for the right side.

I could give u the only hint possible,but that would be useless.So i'll better show u my version.Logarithmate both sides of the equation to find:
[tex]x\ln 2=\ln 7+x\ln 5 =>x=\frac{\ln 7}{\ln 2-\ln 5}[/tex].

Okay...??Clear?

PS:Shmoe,u were quicker than me...Again. :-p
 
Yeah thanks a lot i was messing up and having ln7*xln5
 

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