Question on solving an equation involving logs

[Pyroadept]

Homework Statement

(Here, by 'log' I mean natural logarithm)

Solve for x:

a.log(x) + b.log(1-x) = c

for a, b and c constants

Homework Equations

The Attempt at a Solution

Hi everyone,

This is so embarrassing but this is really stumping me! I know how to do it if a=b:

a.log(x) + a.log(1-x) = c

log(x) + log(1-x) = c/a

log(x(1-x)) = c/a

x(1-x) = exp(c/a)

and then you get a quadratic in x which you can solve.

However, I can't work out how to solve it if a is not equal to b, as I can't see how to group up the logs nicely without ending up with horrible powers of x. What direction should I go in?

Thanks for any help!

 

[SammyS]

Homework Statement

(Here, by 'log' I mean natural logarithm)

Solve for x:

a.log(x) + b.log(1-x) = c

for a, b and c constants

Homework Equations

The Attempt at a Solution

Hi everyone,

This is so embarrassing but this is really stumping me! I know how to do it if a=b:

a.log(x) + a.log(1-x) = c

log(x) + log(1-x) = c/a

log(x(1-x)) = c/a

x(1-x) = exp(c/a)

and then you get a quadratic in x which you can solve.

However, I can't work out how to solve it if a is not equal to b, as I can't see how to group up the logs nicely without ending up with horrible powers of x. What direction should I go in?

Thanks for any help!

For this general case, I doubt that any approach will give a "nice" result.

Divide both sides by b.

Use properties of logarithms:

a∙log(u) = log(u^a)

log(p)+log(q) = log(p∙q)

 

[DryRun]

(Here, by 'log' I mean natural logarithm)

Wouldn't it be simpler if you had just typed $\ln$ then?
$$a\ln (x) + b\ln (1-x) = c \\\ln x^a+\ln (1-x)^b=c$$

 

[Pyroadept]

I thought it might come out looking like an 'In'... :)

So if I get it down to:

p$^{a}$(1-p)$^{b}$ = e^c

how can I solve for p then? Assuming the a and b are large-ish numbers like, say, 10 and 20.

 

[DryRun]

I thought it might come out looking like an 'In'... :)

So if I get it down to:

p$^{a}$(1-p)$^{b}$ = e^c

how can I solve for p then? Assuming the a and b are large-ish numbers like, say, 10 and 20.

Since $a\not = b$, i think you'll have to first expand $(1-p)^b$ using binomial theorem, and then multiply $p^a$ into that expansion.

 

[SammyS]

I thought it might come out looking like an 'In'... :)

So if I get it down to:

p$^{a}$(1-p)$^{b}$ = e^c

how can I solve for p then? Assuming the a and b are large-ish numbers like, say, 10 and 20.

Of course you could rewrite this as

$\displaystyle p^{\,a/b}(1-p)=e^{\,c/b}$​

 

[Pyroadept]

Thanks guys, I appreciate it! :)