Logarithm

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Albi

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Guys I'm trying to solve for [MATH]\log_{x}c[/MATH]what should I do?

[MATH]\frac{1}{r}=\frac{1}{\frac{1}{p}+\frac{1}{q}+log_{x}c}[/MATH]
 
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I like the way that Subhotosh stated to take the reciprocal. That is how I would do it. But maybe you think differently so I'll supply another way of thinking about it. If two fractions are equal and they both have the same numerator (or denominator) then the denominators (or numerators) must be equal.

Examples: If 7/x = 7/9 then x=9. If x/4 = (2x+3)/4 then x = 2x+3.
 
Albi said:
Guys I'm trying to solve for [MATH]\log_{x}c[/MATH]what should I do?

[MATH]\frac{1}{r}=\frac{1}{\frac{1}{p}+\frac{1}{q}+log_{x}c}[/MATH]
Click to expand...
Jomo said:
I like the way that Subhotosh stated to take the reciprocal. That is how I would do it. But maybe you think differently so I'll supply another way of thinking about it. If two fractions are equal and they both have the same numerator (or denominator) then the denominators (or numerators) must be equal.

Examples: If 7/x = 7/9 then x=9. If x/4 = (2x+3)/4 then x = 2x+3.
Click to expand...

I tried the Subhotosh way and I got[MATH]r=\frac{q+p+pq*\log_{x}c}{pq}[/MATH] but I'm trying to solve for [MATH]\log_{x}c[/MATH] any idea for the next step?
 
You don't see that

[MATH]\dfrac{1}{r}=\frac{1}{\frac{1}{p}+\frac{1}{q}+log_{x}c} \implies[/MATH]
[MATH]r = log_x(c) + \dfrac{1}{p} + \dfrac{1}{q} \implies[/MATH]
[MATH]log_x(c) = r - \dfrac{1}{p} - \dfrac{1}{q}.[/MATH]
"Come on" seems like a very mild expression under the circumstances. Mine would probably get me banned for weeks.
 
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