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

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The equation 5^(log(3X) – log(3)2) = 125 can be solved by recognizing that 125 equals 5^3. This leads to the equation log(3X) - log(3)2 = 3, which simplifies to log(3X) = 3 + (log2/log3). The correct interpretation of log(3X) as log3x allows for the solution x = 54, confirming the answer provided in the book. The discussion emphasizes the importance of correctly identifying logarithmic bases in solving exponential equations.

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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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