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

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}

9^{l o g_{3} (x)} x^{3} = 243

ln (T) = 1.56 + 0.82 \cdot ln (22)

0 = ln (0 + x)

lo g_{3} (x + 3) + 3 x = 5

2^{l o g_{10} (n)} = n

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}