What is the answer to ln((t-65)/(125))=-0.07t ?

The answer to ln((t-65)/(125))=-0.07t is t=(\W_{0}((35)/(4e^{\frac{31.85)/7)})+4.55}{0.07}

Solve for t: ln((t-65)/(125))=-0.07t

The solution is t=(\W_{0}((35)/(4e^{\frac{31.85)/7)})+4.55}{0.07}

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ln (\frac{t - 65}{125}) = - 0.07 t

t = \frac{W _{0} ( \frac{35}{4 e ^{\frac{31.85}{7}}} ) + 4.55}{0.07}

Solution steps

ln (\frac{t - 65}{125}) = - 0.07 t

f (x) = g (x),

then

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

e^{l n (\frac{t - 65}{125})} = e^{- 0.07 t}

Simplify

e^{l n (\frac{t - 65}{125})} : \frac{t - 65}{125}

\frac{t - 65}{125} = e^{- 0.07 t}

Multiply both sides by

125

\frac{t - 65}{125} \cdot 125 = e^{- 0.07 t} \cdot 125

Simplify

t - 65 = e^{- 0.07 t} \cdot 125

Solve

t - 65 = e^{- 0.07 t} \cdot 125 : t = \frac{W _{0} ( \frac{35}{4 e ^{\frac{31.85}{7}}} ) + 4.55}{0.07}

t = \frac{W _{0} ( \frac{35}{4 e ^{\frac{31.85}{7}}} ) + 4.55}{0.07}

Verify Solutions:

t = \frac{W _{0} ( \frac{35}{4 e ^{\frac{31.85}{7}}} ) + 4.55}{0.07}

True

The solution is

t = \frac{W _{0} ( \frac{35}{4 e ^{\frac{31.85}{7}}} ) + 4.55}{0.07}

lo g_{a} (x) = lo g_{a} (x)

y, ln (y) = 4 ln (x) + c

lo g_{10} ((x + 2) (x - 1)) = 1

1.73 = - lo g_{10} (x)

10 0^{l o g_{10} (x)} = 2 x^{2}

What is the answer to ln((t-65)/(125))=-0.07t ?
The answer to ln((t-65)/(125))=-0.07t is t=(\W_{0}((35)/(4e^{\frac{31.85)/7)})+4.55}{0.07}
Solve for t: ln((t-65)/(125))=-0.07t
The solution is t=(\W_{0}((35)/(4e^{\frac{31.85)/7)})+4.55}{0.07}