What is ln(10)-ln(10^{-11}e^{40x})-ln(y) ?

The solution to ln(10)-ln(10^{-11}e^{40x})-ln(y) is ln((10^{12})/(e^{40x)y})

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simplify ln(10)-ln(10^{-11}e^{40x})-ln(y)

simplify

ln (10) - ln (1 0^{- 11} e^{40 x}) - ln (y)

ln (\frac{1 0 ^{12}}{e ^{40 x} y})

Solution steps

ln (10) - ln (1 0^{- 11} e^{40 x}) - ln (y)

Apply log rule:

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

ln (10) - ln (1 0^{- 11} e^{40 x}) = ln (\frac{10}{1 0 ^{- 11} e ^{40 x}})

= ln (\frac{10}{1 0 ^{- 11} e ^{40 x}}) - ln (y)

Cancel the common factor:

1 0^{- 11}

= ln (\frac{1 0 ^{12}}{e ^{40 x}}) - ln (y)

Apply log rule:

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

ln (\frac{1 0 ^{12}}{e ^{40 x}}) - ln (y) = ln (\frac{\frac{1 0 ^{12}}{e ^{40 x}}}{y})

= ln (\frac{\frac{1 0 ^{12}}{e ^{40 x}}}{y})

Apply the fraction rule:

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

= ln (\frac{1 0 ^{12}}{e ^{40 x} y})

7 lo g_{e} (x) - 0.5 lo g_{e} (y)

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

ln (4 e^{- \frac{6}{100} t})

20 e^{9} + ln (12) + 50 - ln (9)

- lo g_{10} (2.3 \cdot 1 0^{- 9})

What is ln(10)-ln(10^{-11}e^{40x})-ln(y) ?
The solution to ln(10)-ln(10^{-11}e^{40x})-ln(y) is ln((10^{12})/(e^{40x)y})