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solving ln(x)/ln(x/y)=e^7

lkcl

New member
Feb 8, 2014
7
dear all,

my apologies if this is the wrong location to post, i am a software engineer whose maths was at A and S-Level back in 1987. i am on the exciting but empirical (iterative) track of the electron's magnetic moment: using a program called ries i have an *exact* matching equation - details are here if anyone'e interested -
Physics Discussion Forum ? View topic - [RFC] mass of electron to measured accuracy to date

the problem is that ries generates its solutions in LHS and RHS randomly, LHS is in terms of x, RHS is not. the equation is:

ln(x)/ln(x/y)=e^7

or, actually, ries came out with it as:

ln(x^(e^7)) / ln(x) = 1/(e^7)

and i need to solve that. it looks to me like it's "log (in base x) of x itself divided by a constant equals a constant!" which is... crazed :)

is there actually a mathematical solution to this, or would it be necessary to use iteration (again)?

thank you for any help.

l.
 

lkcl

New member
Feb 8, 2014
7
achh, i fear this may have been down to a range error in the double-precision floating point of standard computer maths libraries. the left hand side 0.16^e^7 is veeery small (3e-3000)
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
dear all,

my apologies if this is the wrong location to post, i am a software engineer whose maths was at A and S-Level back in 1987. i am on the exciting but empirical (iterative) track of the electron's magnetic moment: using a program called ries i have an *exact* matching equation - details are here if anyone'e interested -
Physics Discussion Forum ? View topic - [RFC] mass of electron to measured accuracy to date

the problem is that ries generates its solutions in LHS and RHS randomly, LHS is in terms of x, RHS is not. the equation is:

ln(x)/ln(x/y)=e^7

or, actually, ries came out with it as:

ln(x^(e^7)) / ln(x) = 1/(e^7)

and i need to solve that. it looks to me like it's "log (in base x) of x itself divided by a constant equals a constant!" which is... crazed :)

is there actually a mathematical solution to this, or would it be necessary to use iteration (again)?

thank you for any help.

l.
$\displaystyle \begin{align*} \frac{\ln{(x)}}{\ln{ \left( \frac{x}{y} \right) }} &= e^7 \\ \frac{\ln{(x)}}{\ln{(x)} - \ln{(y)}} &= e^7 \\ \ln{(x)} &= e^7 \left[ \ln{(x)} - \ln{(y)} \right] \\ \ln{(x)} &= e^7\ln{(x)} - e^7\ln{(y)} \\ e^7\ln{(y)} &= e^7\ln{(x)} - \ln{(x)} \\ e^7\ln{(y)} &= \left( e^7 - 1 \right) \ln{(x)} \\ \ln{(y)} &= \left( \frac{e^7 - 1}{e^7} \right) \ln{(x)} \\ \ln{(y)} &= \left( 1 - e^{-7} \right) \ln{(x)} \\ \ln{(y)} &= \ln{ \left( x ^{1 - e^{-7}} \right) } \\ y &= x^{ 1 - e^{-7}} \end{align*}$
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Hello lkcl and welcome to MHB! :D

I have moved this thread as it's really a problem in elementary rather than linear or abstract algebra.

Prove It has provided a solution to the first equation, for $y$ in terms of $x$. Now the second equation you gave is not an identity as it implies \(\displaystyle e^7=e^{-7}\).

To show this consider:

\(\displaystyle \frac{\ln\left(x^{e^7} \right)}{\ln(x)}=\frac{1}{e^7}\)

Now, to the numerator on the left, we may apply the logarithmic property:

\(\displaystyle \log_a\left(b^c \right)=c\cdot\log_a(b)\) to obtain:

\(\displaystyle \frac{e^7\ln(x)}{\ln(x)}=\frac{1}{e^7}\)

Dividing out common factors on the left, we have:

\(\displaystyle e^7=\frac{1}{e^7}\)

And we may rewrite the right side using a negative exponent:

\(\displaystyle e^7=e^{-7}\)

Thus, the given equation is not true, however, the following are true:

\(\displaystyle \frac{\ln\left(x^{e^7} \right)}{\ln(x)}=e^7\)

or

\(\displaystyle \frac{\ln\left(x^{e^{-7}} \right)}{\ln(x)}=\frac{1}{e^7}\)

This is a result of:

\(\displaystyle y=\frac{\ln\left(x^{y} \right)}{\ln(x)}\)

We could also view this using the change of base formula \(\displaystyle \frac{\log_a(b)}{\log_a(c)}=\log_c(b)\) to write:

\(\displaystyle \frac{\ln\left(x^{y} \right)}{\ln(x)}=\log_x\left(x^y \right)=y\log_x(x)=y\cdot1=y\)
 

lkcl

New member
Feb 8, 2014
7
hallo, hallo, thanks for the welcome - yes this is one of those things that i am floundering around with, taking on far more than i can chew, but patiently making progess. ries - amazing program RIES - Find Algebraic Equations, Given Their Solution at MROB - has a few bugs. although its developer has clearly sorted out various bugs removing things like "1=1" or "x/x", finding things like ln(x)/ln(x/y)=y as a "solution" are much harder to.. to... not come up with. as this was the first time i used ries, i didn't know it could make that kind of mistake.

but, here's the good news: thanks to the two posts, i learned about how to manipulate algebraic logarithm equations :) for that i am extremely grateful.

p.s. "Living in the pools, They soon forget about the sea...— Rush, "Natural Science" (1980)" - great album! love jacob's ladder.