Solve 5^(log(3X) – log(3)2) = 125: Find X

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Homework Help Overview

The discussion revolves around solving the equation 5^(log(3X) – log(3)2) = 125, which involves logarithmic properties and exponentiation. Participants are exploring the relationship between logarithmic expressions and their bases.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss rewriting 125 as 5^3 and manipulating logarithmic expressions. There is confusion regarding the interpretation of log(3X) and whether it should be read as log3x. Some suggest exponentiating both sides of the equation as a potential step forward.

Discussion Status

There is an ongoing exploration of the problem with various interpretations being considered. Some participants have provided hints and suggestions for moving forward, but no consensus has been reached on the correct approach or interpretation of the logarithmic terms.

Contextual Notes

Participants are questioning the accuracy of the problem statement as it appears in the book, suggesting there may be a typo affecting their understanding of the logarithmic expressions involved.

punjabi_monster
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math help please!

here is the question i am having trouble with.

5^(log(3X) – log(3)2) = 125...where the base is in brackets.

This is how i attempted to solve for x, but i got stuck. Can someone please help me out. thanks

I understand u can make the 125=5^3 and then:
log(3X) - log(3)2 = 3
log(3X) = 3 + [(log2)/(log3)]
log(3X) = 3.63
Now what? :rolleyes:

The answer in the book is 54.
 
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punjabi_monster said:
here is the question i am having trouble with.

5^(log(3X) – log(3)2) = 125...where the base is in brackets.

This is how i attempted to solve for x, but i got stuck. Can someone please help me out. thanks

I understand u can make the 125=5^3 and then:
log(3X) - log(3)2 = 3
log(3X) = 3 + [(log2)/(log3)]
log(3X) = 3.63
Now what? :rolleyes:

The answer in the book is 54.

Exponentiate both terms of the last equation.Your answer would ly a mere division away.

Daniel.
 
I'm assuming that log(3X) is actually log3x. Recall the definition of the logarithm, that logba = c means that bc = a and vice-versa.
 
no 3x is the base
 
dextercioby said:
Exponentiate both terms of the last equation.Your answer would ly a mere division away.

Daniel.

i don't understand what you are trying to say here :redface:
 
punjabi_monster said:
no 3x is the base
Then what are you taking the logarithm in base 3x of ?
Ie. log3x4 = 12 would be a valid equation (read "the logarithm in base 3x of 4 is 12"), but log3x = 12 is a meaningless fragment.
 
hmmm maybe its a typo in my book.
 
punjabi_monster said:
hmmm maybe its a typo in my book.
Possibly. For the record, reading it as log3x does give x=54.
:smile:
 
yes that makes more sense
 
  • #10
thanks for ur help
 
  • #11
punjabi_monster said:
here is the question i am having trouble with.

5^(log(3X) – log(3)2) = 125...where the base is in brackets.

This is how i attempted to solve for x, but i got stuck. Can someone please help me out. thanks

I understand u can make the 125=5^3 and then:
log(3X) - log(3)2 = 3
log(3X) = 3 + [(log2)/(log3)]
log(3X) = 3.63
Now what? :rolleyes:

The answer in the book is 54.

I think the equation should be:
log(3)x-log(3)2=3
log(3)x=3+log(3)2


log(3)x means logarithm of x with base 3.

At this point you can solve brute force but you can do this without any calculator
Hint: rewrite 3 as a log(3)something then the answer pops right out.
 

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