What is the answer to 1/(h(x))=9^{kx} ?

The answer to 1/(h(x))=9^{kx} is x=(\W((2ln(3)k)/h))/(2ln(3)k)

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1/(h(x))=9^{kx}

\frac{1}{h ( x )} = 9^{k x}

x = \frac{W ( \frac{2 l n ( 3 ) k}{h} )}{2 ln ( 3 ) k}

Solution steps

\frac{1}{h ( x )} = 9^{k x}

Prepare

\frac{1}{h ( x )} = 9^{k x}

for Lambert form:

1 = e^{2 l n (3) k x} h x

Rewrite the equation with

2 ln (3) k x = u

and

x = \frac{u}{2 ln ( 3 ) k}

1 = e^{u} h (\frac{u}{2 ln ( 3 ) k})

Rewrite

1 = e^{u} h (\frac{u}{2 ln ( 3 ) k})

in Lambert form:

e^{u} u = \frac{2 ln ( 3 ) k}{h}

Solve

e^{u} u = \frac{2 ln ( 3 ) k}{h} : u = W (\frac{2 ln ( 3 ) k}{h})

u = W (\frac{2 ln ( 3 ) k}{h})

Substitute back

u = 2 ln (3) k x,

solve for

x

Solve

2 ln (3) k x = W (\frac{2 ln ( 3 ) k}{h}) : x = \frac{W ( \frac{2 l n ( 3 ) k}{h} )}{2 ln ( 3 ) k}

x = \frac{W ( \frac{2 l n ( 3 ) k}{h} )}{2 ln ( 3 ) k}

x, e^{2 x} - 5 e^{x} + 6 = 0

(1 + \frac{0.1}{12})^{12 t} = 2

4^{x - 7} = 2

3^{x} = 4.6

8 \cdot 2^{2 x} + 7 \cdot 2^{x} - 1 = 0

What is the answer to 1/(h(x))=9^{kx} ?
The answer to 1/(h(x))=9^{kx} is x=(\W((2ln(3)k)/h))/(2ln(3)k)