What is the answer to 2^{x-1}=3x ?

The answer to 2^{x-1}=3x is x=-(\W_{-1}(-(ln(2))/6))/(ln(2)),x=-(\W_{0}(-(ln(2))/6))/(ln(2))

Solve for x: 2^{x-1}=3x

The solution is x=-(\W_{-1}(-(ln(2))/6))/(ln(2)),x=-(\W_{0}(-(ln(2))/6))/(ln(2))

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2^{x - 1} = 3 x

x = - \frac{W _{- 1} ( - \frac{l n ( 2 )}{6} )}{ln ( 2 )}, x = - \frac{W _{0} ( - \frac{l n ( 2 )}{6} )}{ln ( 2 )}

Decimal Notation

x = 4.86841 \dots, x = 0.19014 \dots

Solution steps

2^{x - 1} = 3 x

Prepare

2^{x - 1} = 3 x

for Lambert form:

1 = 3 x e^{l n (2) (- x + 1)}

Rewrite the equation with

- x ln (2) = u

and

x = - \frac{u}{ln ( 2 )}

1 = 3 (- \frac{u}{ln ( 2 )}) e^{l n (2) (- (- \frac{u}{l n ( 2 )}) + 1)}

Simplify

1 = - \frac{3 e ^{u + l n (2)} u}{ln ( 2 )}

Rewrite

1 = - \frac{3 e ^{u + l n (2)} u}{ln ( 2 )}

in Lambert form:

e^{u} u = - \frac{ln ( 2 )}{6}

Solve

e^{u} u = - \frac{ln ( 2 )}{6} : u = W_{- 1} (- \frac{ln ( 2 )}{6}), u = W_{0} (- \frac{ln ( 2 )}{6})

u = W_{- 1} (- \frac{ln ( 2 )}{6}), u = W_{0} (- \frac{ln ( 2 )}{6})

Substitute back

u = - x ln (2),

solve for

x

Solve

- x ln (2) = W_{- 1} (- \frac{ln ( 2 )}{6}) : x = - \frac{W _{- 1} ( - \frac{l n ( 2 )}{6} )}{ln ( 2 )}

Solve

- x ln (2) = W_{0} (- \frac{ln ( 2 )}{6}) : x = - \frac{W _{0} ( - \frac{l n ( 2 )}{6} )}{ln ( 2 )}

x = - \frac{W _{- 1} ( - \frac{l n ( 2 )}{6} )}{ln ( 2 )}, x = - \frac{W _{0} ( - \frac{l n ( 2 )}{6} )}{ln ( 2 )}

2^{- x} = 0

e^{8 - 4 x} = 9

e^{8 - 4 x} = 4

1 0^{x} = \frac{1}{100}

5^{x - 2} = 625

What is the answer to 2^{x-1}=3x ?
The answer to 2^{x-1}=3x is x=-(\W_{-1}(-(ln(2))/6))/(ln(2)),x=-(\W_{0}(-(ln(2))/6))/(ln(2))
Solve for x: 2^{x-1}=3x
The solution is x=-(\W_{-1}(-(ln(2))/6))/(ln(2)),x=-(\W_{0}(-(ln(2))/6))/(ln(2))