What is the answer to -11+(12)/T =-4ln(T) ?

The answer to -11+(12)/T =-4ln(T) is T=-3/(\W_{-1)(-3/(e^{11/4)})},T=-3/(\W_{0)(-3/(e^{11/4)})}

Solve for T: -11+(12)/T =-4ln(T)

The solution is T=-3/(\W_{-1)(-3/(e^{11/4)})},T=-3/(\W_{0)(-3/(e^{11/4)})}

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- 11 + \frac{12}{T} = - 4 ln (T)

T = - \frac{3}{W _{- 1} ( - \frac{3}{e ^{\frac{11}{4}}} )}, T = - \frac{3}{W _{0} ( - \frac{3}{e ^{\frac{11}{4}}} )}

Solution steps

- 11 + \frac{12}{T} = - 4 ln (T)

Switch sides

- 4 ln (T) = - 11 + \frac{12}{T}

Divide both sides by

- 4

ln (T) = - \frac{- 11 T + 12}{4 T}

f (x) = g (x),

then

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

e^{l n (T)} = e^{- \frac{- 11 T + 12}{4 T}}

Simplify

e^{l n (T)} : T

T = e^{- \frac{- 11 T + 12}{4 T}}

Solve

T = e^{- \frac{- 11 T + 12}{4 T}} : T = - \frac{3}{W _{- 1} ( - \frac{3}{e ^{\frac{11}{4}}} )}, T = - \frac{3}{W _{0} ( - \frac{3}{e ^{\frac{11}{4}}} )}

T = - \frac{3}{W _{- 1} ( - \frac{3}{e ^{\frac{11}{4}}} )}, T = - \frac{3}{W _{0} ( - \frac{3}{e ^{\frac{11}{4}}} )}

Verify Solutions:

T = - \frac{3}{W _{- 1} ( - \frac{3}{e ^{\frac{11}{4}}} )}

True

, T = - \frac{3}{W _{0} ( - \frac{3}{e ^{\frac{11}{4}}} )}

True

The solutions are

T = - \frac{3}{W _{- 1} ( - \frac{3}{e ^{\frac{11}{4}}} )}, T = - \frac{3}{W _{0} ( - \frac{3}{e ^{\frac{11}{4}}} )}

lo g_{10} (75) = lo g_{10} (11 x)

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

ln ((x - 4)^{2}) = 3.219

lo g_{10} (9) (a) = lo g_{3} (x)

ln (3 w) = ln (4 w - 14)

What is the answer to -11+(12)/T =-4ln(T) ?
The answer to -11+(12)/T =-4ln(T) is T=-3/(\W_{-1)(-3/(e^{11/4)})},T=-3/(\W_{0)(-3/(e^{11/4)})}
Solve for T: -11+(12)/T =-4ln(T)
The solution is T=-3/(\W_{-1)(-3/(e^{11/4)})},T=-3/(\W_{0)(-3/(e^{11/4)})}