Why do logarithms produce such difficult problems This was inspired by Fun logarithm question, beca

Adelyn Rodriguez

Adelyn Rodriguez

Answered question

2022-05-14

Why do logarithms produce such difficult problems
This was inspired by Fun logarithm question, because it made me remember a question I accidentally asked on a quiz some time ago. It was suppose to have both log bases the same, 3 or 5.
log 5 ( x + 3 ) = 1 log 3 ( x 1 )
After apologizing to my students, I talked to some people about it and we could not find an analytical solution... other than to realize that x = 2 is a solution by just trying it. So, my questions are: Is there an analytical solution to this specific problem? And, more importantly, why do variables in exponents/logarithms that are seemingly easy to state produce such difficult problems? I would like some insight into the second question more than the first, as answering the second will also answer the first, I think.

Answer & Explanation

Adeline Shah

Adeline Shah

Beginner2022-05-15Added 18 answers

If you put everything in a common base, let's say 5, this equation is equivalent to
( x + 3 ) ( x 1 ) c = 5
, with c = log 5 3 0.68261. Solving expressions with polynomials is usually easy. I would guess that the problem here is that the exponents are not integer, but real, which makes everything harder.
et1wdwk4o

et1wdwk4o

Beginner2022-05-16Added 2 answers

Also not really an analytical solution, but another way to "visualize" the solution is to make a substitution
3 t = x 1
then the equation transforms to 3 t + 4 = 5.5 t or
3 t 5 t + 4.5 t = 5
which is the same as
( 4 1 ) t ( 4 + 1 ) t + 4 ( 4 + 1 ) t = 1 + 4
Then it is clear that if ( 4 1 ) t ( 4 + 1 ) t = 1 and ( 4 + 1 ) t = 1 then we have solution. Those two equations hold true for t = 0 and plugging it back to the original substitution, we obtain x = 2

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