What is 1/(1/25 ln(2/3))ln((16-20)/(5-20)) ?

The solution to 1/(1/25 ln(2/3))ln((16-20)/(5-20)) is (25(ln(-4)-ln(-15)))/(ln(2/3))

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simplify 1/(1/25 ln(2/3))ln((16-20)/(5-20))

simplify

\frac{1}{\frac{1}{25} ln ( \frac{2}{3} )} ln (\frac{16 - 20}{5 - 20})

\frac{25 ( ln ( - 4 ) - ln ( - 15 ) )}{ln ( \frac{2}{3} )}

Solution steps

\frac{1}{\frac{1}{25} ln ( \frac{2}{3} )} ln (\frac{16 - 20}{5 - 20})

Apply log rule:

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

ln (\frac{16 - 20}{5 - 20}) = ln (16 - 20) - ln (5 - 20)

= \frac{1}{\frac{1}{25} ln ( \frac{2}{3} )} (ln (16 - 20) - ln (5 - 20))

\frac{1}{\frac{1}{25} ln ( \frac{2}{3} )} = \frac{25}{ln ( \frac{2}{3} )}

= \frac{25}{ln ( \frac{2}{3} )} (ln (16 - 20) - ln (5 - 20))

Subtract the numbers:

16 - 20 = - 4

= \frac{25}{ln ( \frac{2}{3} )} (ln (- 4) - ln (5 - 20))

Subtract the numbers:

5 - 20 = - 15

= \frac{25}{ln ( \frac{2}{3} )} (ln (- 4) - ln (- 15))

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{25 ( ln ( - 4 ) - ln ( - 15 ) )}{ln ( \frac{2}{3} )}

\frac{5 π}{8} ln (5)

- 0.7 lo g_{10} (20.7) - 0.3 lo g_{10} (20.3)

6 lo g_{10} (x) - lo g_{10} (17)

\frac{1}{0.02} ln (35 (e^{0.01} - e^{- 0.02}))

ln (\frac{( 1 - 1.25 )}{0.075}) - 1

What is 1/(1/25 ln(2/3))ln((16-20)/(5-20)) ?
The solution to 1/(1/25 ln(2/3))ln((16-20)/(5-20)) is (25(ln(-4)-ln(-15)))/(ln(2/3))