Equation with logarithmic and polynomial terms

AI Thread Summary
The discussion revolves around solving the equation involving both polynomial and logarithmic terms: (x-1)^2 / ((x-1) - 6 log(x-1)) = 3 × 6^(3 log(2) + 2 log(3)). The original poster simplifies the equation to y^2 - 216y - 216 log_6(y) = 0, where y = x - 1. It is noted that the equation cannot be solved using elementary functions and may require the Lambert W function for a solution. There is a suggestion that a simpler solution might exist, given the context of an old high school exam. The conversation highlights the complexities of combining logarithmic and polynomial expressions in equations.
NanakiXIII
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This is not actually a homework question, but it seemed appropriate to put it here. In an old exam from 1921 I found the following problem. I never learned how to solve this type of thing and I haven't been able to figure it out, so: how does one solve this?

Homework Statement



Solve for x:

\frac{(x-1)^2}{(x-1)-^6\log (x-1)} = 3 \times 6^{3\times^6\log 2 + 2\times^6\log 3}

Homework Equations





The Attempt at a Solution



I went ahead and simplified this to

y^2 -72y + 72 ^6\log y=0

where y=x-1, but, as I said, I never learned how to solve this type of equation involving both polynomial terms and logarithms and I don't know how to proceed.
 
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NanakiXIII said:
This is not actually a homework question, but it seemed appropriate to put it here. In an old exam from 1921 I found the following problem. I never learned how to solve this type of thing and I haven't been able to figure it out, so: how does one solve this?

Homework Statement



Solve for x:

\frac{(x-1)^2}{(x-1)-^6\log (x-1)} = 3 \times 6^{3\times^6\log 2 + 2\times^6\log 3}
You have several instances of expressions such as 6log (something). I suspect that these really mean log6(something). IOW, the log expressions are log-base 6. Please clarify.
NanakiXIII said:

Homework Equations





The Attempt at a Solution



I went ahead and simplified this to

y^2 -72y + 72 ^6\log y=0

where y=x-1, but, as I said, I never learned how to solve this type of equation involving both polynomial terms and logarithms and I don't know how to proceed.
 
Yes, exactly. That's how we wrote log-base-6 in school.
 
On the right, use the laws of logarithms: a log(x)= log(x^a), log(x)+ log(y)= log(xy), and b^{log_b(x)}= x to get 3(6^{3log_6(2)+ 2log_6(3)})= 3(6^{log_6((2^3)(3^2)})= 3(8)(9)= 216 By the way, it is NOT a good idea to use "\times" to indicate multiplication when you have x as the unknown. Just use parentheses.
 
Yes, it would appear I miscalculated, the 72's in my equation should be 216's. I'll edit my post. That said, though, I still don't know how to solve the equation.

Edit: Actually, it appears I can't edit the original post, so here's an erratum:

My simplification (last equation) should be

y^2 - 216 y - 216 \log_6 y = 0

where y = x-1.
 
Last edited:
Since that equation involves both powers of y and logarithm of y, the solution cannot be written in terms of "elementary" functions. It should be possible, by taking the exponential of both sides, to get it in the form ve^v= \text{constant} and then solve it in terms of the Lambert W function.
 
Well, that explains why I never learned to solve this. However, since this is an old high school exam, I would be surprised if there was not a simpler solution. It is probably possible to solve it using elementary functions and a logarithm table.
 
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