What is the answer to 4^{-6}x=17^x+4 ?

The answer to 4^{-6}x=17^x+4 is x=-(\W(-17^{16384}*4096ln(17))-16384ln(17))/(ln(17))

Solve for x: 4^{-6}x=17^x+4

The solution is x=-(\W(-17^{16384}*4096ln(17))-16384ln(17))/(ln(17))

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4^{- 6} x = 1 7^{x} + 4

x = - \frac{W ( - 1 7 ^{16384} \cdot 4096 ln ( 17 ) ) - 16384 ln ( 17 )}{ln ( 17 )}

Solution steps

4^{- 6} x = 1 7^{x} + 4

Prepare

4^{- 6} x = 1 7^{x} + 4

for Lambert form:

1 = e^{- l n (17) x} (\frac{1}{4096} x - 4)

Rewrite the equation with

(- x + 16384) ln (17) = u

and

x = - \frac{u - 16384 ln ( 17 )}{ln ( 17 )}

1 = e^{- l n (17) (- \frac{u - 16384 l n ( 17 )}{l n ( 17 )})} (\frac{1}{4096} (- \frac{u - 16384 ln ( 17 )}{ln ( 17 )}) - 4)

Simplify

1 = e^{u - 16384 l n (17)} (- \frac{u - 16384 ln ( 17 )}{4096 ln ( 17 )} - 4)

Rewrite

1 = e^{u - 16384 l n (17)} (- \frac{u - 16384 ln ( 17 )}{4096 ln ( 17 )} - 4)

in Lambert form:

e^{u} u = - 1 7^{16384} \cdot 4096 ln (17)

Solve

e^{u} u = - 1 7^{16384} \cdot 4096 ln (17) : u = W (- 1 7^{16384} \cdot 4096 ln (17))

u = W (- 1 7^{16384} \cdot 4096 ln (17))

Substitute back

u = (- x + 16384) ln (17),

solve for

x

Solve

(- x + 16384) ln (17) = W (- 1 7^{16384} \cdot 4096 ln (17)) : x = - \frac{W ( - 1 7 ^{16384} \cdot 4096 ln ( 17 ) ) - 16384 ln ( 17 )}{ln ( 17 )}

x = - \frac{W ( - 1 7 ^{16384} \cdot 4096 ln ( 17 ) ) - 16384 ln ( 17 )}{ln ( 17 )}

4^{x} \cdot 4^{2 x - 8} = 64

(\frac{1}{4})^{3 - x} = 64

2 5^{x} + 125 = 5^{x + 1}

100 0^{1000} = 10 0^{x}

\frac{3 ^{x}}{2} = 27

What is the answer to 4^{-6}x=17^x+4 ?
The answer to 4^{-6}x=17^x+4 is x=-(\W(-17^{16384}*4096ln(17))-16384ln(17))/(ln(17))
Solve for x: 4^{-6}x=17^x+4
The solution is x=-(\W(-17^{16384}*4096ln(17))-16384ln(17))/(ln(17))