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Popular >

27^3x-1= 1/(81^x)

Solution

2 7^{3} x - 1 = \frac{1}{8 1 ^{x}}

Solution

x = \frac{W _{0} ( \frac{4 l n ( 3 )}{19683 e ^{\frac{4 l n ( 3 )}{19683}}} )}{4 ln ( 3 )} + \frac{1}{19683}

+1

Decimal Notation

x = 0.00010 \dots

Solution steps

2 7^{3} x - 1 = \frac{1}{8 1 ^{x}}

Prepare

2 7^{3} x - 1 = \frac{1}{8 1 ^{x}}

for Lambert form:

e^{4 l n (3) x} (19683 x - 1) = 1

Rewrite the equation with

4 ln (3) (x - \frac{1}{19683}) = u

and

x = \frac{u}{4 ln ( 3 )} + \frac{1}{19683}

e^{4 l n (3) (\frac{u}{4 l n ( 3 )} + \frac{1}{19683})} (19683 (\frac{u}{4 ln ( 3 )} + \frac{1}{19683}) - 1) = 1

Simplify

\frac{19683 e ^{u + \frac{4}{19683} l n (3)} u}{4 ln ( 3 )} = 1

Rewrite

\frac{19683 e ^{u + \frac{4}{19683} l n (3)} u}{4 ln ( 3 )} = 1

in Lambert form:

\frac{e ^{\frac{19683 u + 4 l n ( 3 )}{19683}} u}{e ^{\frac{4 l n ( 3 )}{19683}}} = \frac{4 ln ( 3 )}{19683 e ^{\frac{4 l n ( 3 )}{19683}}}

Solve

\frac{e ^{\frac{19683 u + 4 l n ( 3 )}{19683}} u}{e ^{\frac{4 l n ( 3 )}{19683}}} = \frac{4 ln ( 3 )}{19683 e ^{\frac{4 l n ( 3 )}{19683}}} : u = W_{0} (\frac{4 ln ( 3 )}{19683 e ^{\frac{4 l n ( 3 )}{19683}}})

u = W_{0} (\frac{4 ln ( 3 )}{19683 e ^{\frac{4 l n ( 3 )}{19683}}})

Substitute back

u = 4 ln (3) (x - \frac{1}{19683}),

solve for

x

Solve

4 ln (3) (x - \frac{1}{19683}) = W_{0} (\frac{4 ln ( 3 )}{19683 e ^{\frac{4 l n ( 3 )}{19683}}}) : x = \frac{W _{0} ( \frac{4 l n ( 3 )}{19683 e ^{\frac{4 l n ( 3 )}{19683}}} )}{4 ln ( 3 )} + \frac{1}{19683}

x = \frac{W _{0} ( \frac{4 l n ( 3 )}{19683 e ^{\frac{4 l n ( 3 )}{19683}}} )}{4 ln ( 3 )} + \frac{1}{19683}

Verify Solutions:

x = \frac{W _{0} ( \frac{4 l n ( 3 )}{19683 e ^{\frac{4 l n ( 3 )}{19683}}} )}{4 ln ( 3 )} + \frac{1}{19683}

True

The solution is

x = \frac{W _{0} ( \frac{4 l n ( 3 )}{19683 e ^{\frac{4 l n ( 3 )}{19683}}} )}{4 ln ( 3 )} + \frac{1}{19683}

Verify Solutions:

x = \frac{W _{0} ( \frac{4 l n ( 3 )}{19683 e ^{\frac{4 l n ( 3 )}{19683}}} )}{4 ln ( 3 )} + \frac{1}{19683}

True

The solution is

x = \frac{W _{0} ( \frac{4 l n ( 3 )}{19683 e ^{\frac{4 l n ( 3 )}{19683}}} )}{4 ln ( 3 )} + \frac{1}{19683}

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