What is (ln(x+2)-ln(x+1))/(ln(x+1)-ln(x+0)) ?

The solution to (ln(x+2)-ln(x+1))/(ln(x+1)-ln(x+0)) is (ln((x+2)/(x+1))}{ln(\frac{x+1)/x)}

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simplify (ln(x+2)-ln(x+1))/(ln(x+1)-ln(x+0))

simplify

\frac{ln ( x + 2 ) - ln ( x + 1 )}{ln ( x + 1 ) - ln ( x + 0 )}

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

Solution steps

\frac{ln ( x + 2 ) - ln ( x + 1 )}{ln ( x + 1 ) - ln ( x + 0 )}

Apply log rule:

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

ln (x + 2) - ln (x + 1) = ln (\frac{x + 2}{x + 1})

= \frac{ln ( \frac{x + 2}{x + 1} )}{ln ( x + 1 ) - ln ( x + 0 )}

Apply log rule:

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

ln (x + 1) - ln (x + 0) = ln (\frac{x + 1}{x + 0})

= \frac{ln ( \frac{x + 2}{x + 1} )}{ln ( \frac{x + 1}{x + 0} )}

x + 0 = x

= \frac{ln ( \frac{x + 2}{x + 1} )}{ln ( \frac{x + 1}{x} )}

\frac{ln ( \frac{1}{2} )}{ln ( 364 ) - ln ( 365 )}

lo g_{10} (\frac{x z}{Y})

- lo g_{10} (4.9 \cdot 1 0^{- 6})

4 ln ∣ x - 1 ∣ + 6 ln ∣ x ∣ - \frac{1}{x} + C

4 lo g_{10} (x) - \frac{1}{3} lo g_{10} (x + 6)

What is (ln(x+2)-ln(x+1))/(ln(x+1)-ln(x+0)) ?
The solution to (ln(x+2)-ln(x+1))/(ln(x+1)-ln(x+0)) is (ln((x+2)/(x+1))}{ln(\frac{x+1)/x)}