Solving 5^(log₃(3x) – log₃(2)) = 125 for x

[punjabi_monster]

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?

The answer in the book is 54.

 

[dextercioby]

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?

The answer in the book is 54.

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

Daniel.

 

[hypermorphism]

I'm assuming that log(3X) is actually log₃x. Recall the definition of the logarithm, that log_ba = c means that b^c = a and vice-versa.

 

[punjabi_monster]

no 3x is the base

 

[punjabi_monster]

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

 

[hypermorphism]

no 3x is the base

Then what are you taking the logarithm in base 3x of ?
Ie. log_3x4 = 12 would be a valid equation (read "the logarithm in base 3x of 4 is 12"), but log_3x = 12 is a meaningless fragment.

 

[punjabi_monster]

hmmm maybe its a typo in my book.

 

[hypermorphism]

hmmm maybe its a typo in my book.

Possibly. For the record, reading it as log₃x does give x=54.

 

[punjabi_monster]

yes that makes more sense

 

[punjabi_monster]

thanks for ur help

 

[learningphysics]

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?

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.