What is 1/(ln(s+a)-ln(s+b)) ?

The solution to 1/(ln(s+a)-ln(s+b)) is 1/(ln(\frac{s+a){s+b})}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

simplify 1/(ln(s+a)-ln(s+b))

simplify

\frac{1}{ln ( s + a ) - ln ( s + b )}

\frac{1}{ln ( \frac{s + a}{s + b} )}

Solution steps

\frac{1}{ln ( s + a ) - ln ( s + b )}

Apply log rule:

lo g_{c} (a) - lo g_{c} (b) = lo g_{c} (\frac{a}{b})

ln (s + a) - ln (s + b) = ln (\frac{s + a}{s + b})

= \frac{1}{ln ( \frac{s + a}{s + b} )}

ln (4) + ln (2 x)

ln (5 \cdot e^{2}) - ln (e)

ln (\frac{1}{2} (2 + 8))

lo g_{10} (z^{5} y)

What is 1/(ln(s+a)-ln(s+b)) ?
The solution to 1/(ln(s+a)-ln(s+b)) is 1/(ln(\frac{s+a){s+b})}