What is the answer to alog_{3}(4)=2.4^{-a} ?

The answer to alog_{3}(4)=2.4^{-a} is a=(\W_{0}(\frac{ln(2.4))/(2log_{3)(2)})}{ln(2.4)}

Solve for a: alog_{3}(4)=2.4^{-a}

The solution is a=(\W_{0}(\frac{ln(2.4))/(2log_{3)(2)})}{ln(2.4)}

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a lo g_{3} (4) = 2. 4^{- a}

a = \frac{W _{0} ( \frac{l n ( 2.4 )}{2 l o g _{3} ( 2 )} )}{ln ( 2.4 )}

Decimal Notation

a = 0.50798 \dots

Solution steps

a lo g_{3} (4) = 2. 4^{- a}

Prepare

a lo g_{3} (4) = 2. 4^{- a}

for Lambert form:

2 lo g_{3} (2) e^{l n (2.4) a} a = 1

Rewrite the equation with

a ln (2.4) = u

and

a = \frac{u}{ln ( 2.4 )}

2 lo g_{3} (2) e^{u} (\frac{u}{ln ( 2.4 )}) = 1

Rewrite

2 lo g_{3} (2) e^{u} (\frac{u}{ln ( 2.4 )}) = 1

in Lambert form:

e^{u} u = \frac{ln ( 2.4 )}{2 lo g _{3} ( 2 )}

Solve

e^{u} u = \frac{ln ( 2.4 )}{2 lo g _{3} ( 2 )} : u = W_{0} (\frac{ln ( 2.4 )}{2 lo g _{3} ( 2 )})

u = W_{0} (\frac{ln ( 2.4 )}{2 lo g _{3} ( 2 )})

Substitute back

u = a ln (2.4),

solve for

a

Solve

a ln (2.4) = W_{0} (\frac{ln ( 2.4 )}{2 lo g _{3} ( 2 )}) : a = \frac{W _{0} ( \frac{l n ( 2.4 )}{2 l o g _{3} ( 2 )} )}{ln ( 2.4 )}

a = \frac{W _{0} ( \frac{l n ( 2.4 )}{2 l o g _{3} ( 2 )} )}{ln ( 2.4 )}

7^{(- x + 4)} = 1

0.284 3^{x} = 70

9^{x - 1} = 7

3 e^{x} + 1 = 5

e^{2 x} + e^{3 x} = 0

What is the answer to alog_{3}(4)=2.4^{-a} ?
The answer to alog_{3}(4)=2.4^{-a} is a=(\W_{0}(\frac{ln(2.4))/(2log_{3)(2)})}{ln(2.4)}
Solve for a: alog_{3}(4)=2.4^{-a}
The solution is a=(\W_{0}(\frac{ln(2.4))/(2log_{3)(2)})}{ln(2.4)}