Treat it like a "fraction problem" and get reciprocal of both sides: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]
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]
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.
Come on! \(\log_x(c)=r-\dfrac{1}{p}-\dfrac{1}{q}\)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?
Come on were my exact words when I saw the OPs last post.Come on! \(\log_x(c)=r-\dfrac{1}{p}-\dfrac{1}{q}\)