What is the answer to 8^x+1+8^1-x=20 ?

The answer to 8^x+1+8^1-x=20 is x=-(\W_{0}(-\frac{3ln(2))/(2^{33)})+33ln(2)}{3ln(2)}

Solve for x: 8^x+1+8^1-x=20

The solution is x=-(\W_{0}(-\frac{3ln(2))/(2^{33)})+33ln(2)}{3ln(2)}

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8^{x} + 1 + 8^{1} - x = 20

x = - \frac{W _{0} ( - \frac{3 l n ( 2 )}{2 ^{33}} ) + 33 ln ( 2 )}{3 ln ( 2 )}

Decimal Notation

x = - 10.99999 \dots

Solution steps

8^{x} + 1 + 8^{1} - x = 20

Prepare

8^{x} + 1 + 8^{1} - x = 20

for Lambert form:

1 = (x + 11) e^{- 3 l n (2) x}

Rewrite the equation with

3 ln (2) (- x - 11) = u

and

x = - \frac{u + 33 ln ( 2 )}{3 ln ( 2 )}

1 = ((- \frac{u + 33 ln ( 2 )}{3 ln ( 2 )}) + 11) e^{- 3 l n (2) (- \frac{u + 33 l n ( 2 )}{3 l n ( 2 )})}

Simplify

1 = e^{u + 33 l n (2)} (- \frac{u + 33 ln ( 2 )}{3 ln ( 2 )} + 11)

Rewrite

1 = e^{u + 33 l n (2)} (- \frac{u + 33 ln ( 2 )}{3 ln ( 2 )} + 11)

in Lambert form:

e^{u} u = - \frac{3 ln ( 2 )}{2 ^{33}}

Solve

e^{u} u = - \frac{3 ln ( 2 )}{2 ^{33}} : u = W_{0} (- \frac{3 ln ( 2 )}{2 ^{33}})

u = W_{0} (- \frac{3 ln ( 2 )}{2 ^{33}})

Substitute back

u = 3 ln (2) (- x - 11),

solve for

x

Solve

3 ln (2) (- x - 11) = W_{0} (- \frac{3 ln ( 2 )}{2 ^{33}}) : x = - \frac{W _{0} ( - \frac{3 l n ( 2 )}{2 ^{33}} ) + 33 ln ( 2 )}{3 ln ( 2 )}

x = - \frac{W _{0} ( - \frac{3 l n ( 2 )}{2 ^{33}} ) + 33 ln ( 2 )}{3 ln ( 2 )}

- e^{- \frac{1}{2} x} + 1 = 0.4

(\frac{3}{5})^{x} = 2

7^{x + 6} = 3^{x}

4^{3 x + 5} = 9

7^{x - 1} = e^{x}

What is the answer to 8^x+1+8^1-x=20 ?
The answer to 8^x+1+8^1-x=20 is x=-(\W_{0}(-\frac{3ln(2))/(2^{33)})+33ln(2)}{3ln(2)}
Solve for x: 8^x+1+8^1-x=20
The solution is x=-(\W_{0}(-\frac{3ln(2))/(2^{33)})+33ln(2)}{3ln(2)}