What is the answer to ln(((t-20))/(200))=-0.11t ?

The answer to ln(((t-20))/(200))=-0.11t is t=(\W_{0}(\frac{22)/(e^{\frac{24.2){11}}})+2.2}{0.11}

Solve for t: ln(((t-20))/(200))=-0.11t

The solution is t=(\W_{0}(\frac{22)/(e^{\frac{24.2){11}}})+2.2}{0.11}

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ln (\frac{( t - 20 )}{200}) = - 0.11 t

t = \frac{W _{0} ( \frac{22}{e ^{\frac{24.2}{11}}} ) + 2.2}{0.11}

Decimal Notation

t = 28.60245 \dots

Solution steps

ln (\frac{( t - 20 )}{200}) = - 0.11 t

f (x) = g (x),

then

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

e^{l n (\frac{t - 20}{200})} = e^{- 0.11 t}

Simplify

e^{l n (\frac{t - 20}{200})} : \frac{t - 20}{200}

\frac{t - 20}{200} = e^{- 0.11 t}

Multiply both sides by

200

\frac{t - 20}{200} \cdot 200 = e^{- 0.11 t} \cdot 200

Simplify

t - 20 = e^{- 0.11 t} \cdot 200

Solve

t - 20 = e^{- 0.11 t} \cdot 200 : t = \frac{W _{0} ( \frac{22}{e ^{\frac{24.2}{11}}} ) + 2.2}{0.11}

t = \frac{W _{0} ( \frac{22}{e ^{\frac{24.2}{11}}} ) + 2.2}{0.11}

Verify Solutions:

t = \frac{W _{0} ( \frac{22}{e ^{\frac{24.2}{11}}} ) + 2.2}{0.11}

True

The solution is

t = \frac{W _{0} ( \frac{22}{e ^{\frac{24.2}{11}}} ) + 2.2}{0.11}

lo g_{x} (3) = e^{2}

5 = 1 + 3.32 lo g_{10} (x)

ln (x) = - 2.81 \cdot 600 + 1.7

lo g_{2} (1 + x) = b

4 lo g_{2} (3) = lo g_{2} (x)

What is the answer to ln(((t-20))/(200))=-0.11t ?
The answer to ln(((t-20))/(200))=-0.11t is t=(\W_{0}(\frac{22)/(e^{\frac{24.2){11}}})+2.2}{0.11}
Solve for t: ln(((t-20))/(200))=-0.11t
The solution is t=(\W_{0}(\frac{22)/(e^{\frac{24.2){11}}})+2.2}{0.11}