What is 1/(e^{2(ln|s+100|-2ln(10)))} ?

The solution to 1/(e^{2(ln|s+100|-2ln(10)))} is (10000)/(|s+100|^2)

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simplify 1/(e^{2(ln|s+100|-2ln(10)))}

simplify

\frac{1}{e ^{2 (l n ∣ s + 100 ∣ - 2 l n (10))}}

\frac{10000}{∣ s + 100 ∣ ^{2}}

Solution steps

\frac{1}{e ^{2 (l n ∣ s + 100 ∣ - 2 l n (10))}}

Apply log rule:

a lo g_{c} (b) = lo g_{c} (b^{a})

2 ln (10) = ln (1 0^{2})

= \frac{1}{e ^{2 (l n ∣ s + 100 ∣ - l n (1 0^{2}))}}

Apply log rule:

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

ln ∣ s + 100 ∣ - ln (1 0^{2}) = ln (\frac{∣ s + 100 ∣}{1 0 ^{2}})

= \frac{1}{e ^{2 l n (\frac{∣ s + 100 ∣}{1 0 ^{2}})}}

1 0^{2} = 100

= \frac{1}{e ^{2 l n (\frac{∣ s + 100 ∣}{100})}}

e^{2 l n (\frac{∣ s + 100 ∣}{100})} = \frac{∣ s + 100 ∣ ^{2}}{10 0 ^{2}}

= \frac{1}{\frac{∣ s + 100 ∣ ^{2}}{10 0 ^{2}}}

Apply the fraction rule:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

= \frac{10 0 ^{2}}{∣ s + 100 ∣ ^{2}}

10 0^{2} = 10000

= \frac{10000}{∣ s + 100 ∣ ^{2}}

- ln (\frac{3}{2}) + ln (\frac{5}{3})

ln (\frac{1000}{100})

ln (7 x) - ln (1)

- lo g_{10} (7.21 \cdot 1 0^{- 6})

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

What is 1/(e^{2(ln|s+100|-2ln(10)))} ?
The solution to 1/(e^{2(ln|s+100|-2ln(10)))} is (10000)/(|s+100|^2)