What is |(ln(m+Δm)-ln(m-Δm))/2 | ?

The solution to |(ln(m+Δm)-ln(m-Δm))/2 | is (|ln((1+Δ)/(1-Δ))|)/2

Popular >

simplify |(ln(m+Δm)-ln(m-Δm))/2 |

simplify

\frac{ln ( m + Δ m ) - ln ( m - Δ m )}{2}

\frac{ln ( \frac{1 + Δ}{1 - Δ} )}{2}

Solution steps

\frac{ln ( m + Δ m ) - ln ( m - Δ m )}{2}

Apply log rule:

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

ln (m + Δ m) - ln (m - Δ m) = ln (\frac{m + Δ m}{m - Δ m})

= \frac{ln ( \frac{m + Δ m}{m - Δ m} )}{2}

Simplify

\frac{m + Δ m}{m - Δ m} : \frac{1 + Δ}{1 - Δ}

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

Apply absolute rule:

\frac{a}{b} = \frac{∣ a ∣}{∣ b ∣}

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

Apply absolute rule:

∣ a ∣ = a, a \geq 0

∣ 2 ∣ = 2

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

\frac{1}{2} ln (9 x^{2}) + ln (3 x)

ln (\frac{( 2 x + 3 ) ( x + 6 ) ^{4}}{( 1 - 4 x ) ^{5}})

ln (x^{2} - 1) - ln (x + 1) + ln (x^{2} + 1)

5 lo g_{4} (x) - lo g_{4} (6)

ln (\frac{2 \cdot 1 0 ^{6}}{1 0 ^{6}})

What is |(ln(m+Δm)-ln(m-Δm))/2 | ?
The solution to |(ln(m+Δm)-ln(m-Δm))/2 | is (|ln((1+Δ)/(1-Δ))|)/2