What is the answer to ln(((t-63))/(125))=-0.07t ?

The answer to ln(((t-63))/(125))=-0.07t is t=(\W_{0}((35)/(4e^{\frac{30.87)/7)})+4.41}{0.07}

Solve for t: ln(((t-63))/(125))=-0.07t

The solution is t=(\W_{0}((35)/(4e^{\frac{30.87)/7)})+4.41}{0.07}

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ln (\frac{( t - 63 )}{125}) = - 0.07 t

t = \frac{W _{0} ( \frac{35}{4 e ^{\frac{30.87}{7}}} ) + 4.41}{0.07}

Decimal Notation

t = 64.37953 \dots

Solution steps

ln (\frac{( t - 63 )}{125}) = - 0.07 t

f (x) = g (x),

then

a^{f (x)} = a^{g (x)}

e^{l n (\frac{t - 63}{125})} = e^{- 0.07 t}

Simplify

e^{l n (\frac{t - 63}{125})} : \frac{t - 63}{125}

\frac{t - 63}{125} = e^{- 0.07 t}

Multiply both sides by

125

\frac{t - 63}{125} \cdot 125 = e^{- 0.07 t} \cdot 125

Simplify

t - 63 = e^{- 0.07 t} \cdot 125

Solve

t - 63 = e^{- 0.07 t} \cdot 125 : t = \frac{W _{0} ( \frac{35}{4 e ^{\frac{30.87}{7}}} ) + 4.41}{0.07}

t = \frac{W _{0} ( \frac{35}{4 e ^{\frac{30.87}{7}}} ) + 4.41}{0.07}

Verify Solutions:

t = \frac{W _{0} ( \frac{35}{4 e ^{\frac{30.87}{7}}} ) + 4.41}{0.07}

True

The solution is

t = \frac{W _{0} ( \frac{35}{4 e ^{\frac{30.87}{7}}} ) + 4.41}{0.07}

2 ln (4 x + 1) + 14 = 20

7.5 = - lo g_{10} (h)

x, lo g_{10} (x) = - 0.123

- lo g_{10} (x) = 12

lo g_{2} (x - 3) = 3 - lo g_{2} (x - 1)

What is the answer to ln(((t-63))/(125))=-0.07t ?
The answer to ln(((t-63))/(125))=-0.07t is t=(\W_{0}((35)/(4e^{\frac{30.87)/7)})+4.41}{0.07}
Solve for t: ln(((t-63))/(125))=-0.07t
The solution is t=(\W_{0}((35)/(4e^{\frac{30.87)/7)})+4.41}{0.07}