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

The answer to ln((t-20)/(200))=-0.11 is t=(200)/(e^{0.11)}+20

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

The solution is t=(200)/(e^{0.11)}+20

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

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

Decimal

t = 199.16682 \dots

Solution steps

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

Apply log rules

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

Multiply both sides by

200

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

Simplify

e^{- 0.11} \cdot 200 : \frac{200}{e ^{0.11}}

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

Solve

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

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

Verify Solutions:

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

True

The solution is

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

ln (x) + ln (x^{2}) = 4

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

ln (x) + ln (x^{2}) = 3

2 lo g_{10} (3 - x) = - 1

lo g_{10} (x) 49 = 2

What is the answer to ln((t-20)/(200))=-0.11 ?
The answer to ln((t-20)/(200))=-0.11 is t=(200)/(e^{0.11)}+20
Solve for t: ln((t-20)/(200))=-0.11
The solution is t=(200)/(e^{0.11)}+20