Solve for x: question on logarithms. The question: \log_3 x \cdot \log_4 x \cdot \log_5 x

Molecca89g

Molecca89g

Answered question

2022-04-19

Solve for x: question on logarithms.
The question:
log3xlog4xlog5x=log3xlog4xlog5xlog5xlog4xlog3x
My mother who's a math teacher was asked this by one of her students, and she can't quite figure it out. Anyone got any ideas?

Answer & Explanation

alastrimsmnr

alastrimsmnr

Beginner2022-04-20Added 18 answers

Following up on Jaeyong Chung's answer, and working it out:
1=log3xlog4xlog5x
1=(lnx)3ln3ln4ln5
(lnx)3=ln3ln4ln5
(lnx)=ln3ln4ln53
x=exp(ln3ln4ln53)3.85093
EDIT: And, of course, the obvious answer that everyone will overlook: x=1 makes both sides of the equation zero. :D

Elsaidrge

Elsaidrge

Beginner2022-04-21Added 11 answers

The "change-of-base" formula should at least allow you to reduce this to
log3x(log4xlog43)(log5xlog53) = (log3x)3 = 1log43log53.
The number won't be "pretty"...

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