Solve 1/h*sqrt[x]{6}=sqrt[2x]{6}+3

$$\frac { 1 } { h } \cdot \sqrt[ x ] { 6 } = \sqrt[ 2 x ] { 6 } + 3$$

Solve for h

$$h=\frac{6^{\frac{1}{x}}}{6^{\frac{1}{2x}}+3}$$

$$x\neq 0$$

Steps for Solving Linear Equation

Variable h cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by h.

$$1\sqrt[x]{6}=h\sqrt[2x]{6}+h\times 3$$

Swap sides so that all variable terms are on the left hand side.

$$h\sqrt[2x]{6}+h\times 3=1\sqrt[x]{6}$$

Reorder the terms.

$$h\sqrt[2x]{6}+3h=\sqrt[x]{6}$$

Combine all terms containing h.

$$\left(\sqrt[2x]{6}+3\right)h=\sqrt[x]{6}$$

Divide both sides by \sqrt[2x]{6}+3.

$$\frac{\left(\sqrt[2x]{6}+3\right)h}{\sqrt[2x]{6}+3}=\frac{6^{\frac{1}{x}}}{\sqrt[2x]{6}+3}$$

Dividing by \sqrt[2x]{6}+3 undoes the multiplication by \sqrt[2x]{6}+3.

$$h=\frac{6^{\frac{1}{x}}}{\sqrt[2x]{6}+3}$$

Divide 6^{\frac{1}{x}} by \sqrt[2x]{6}+3.

$$h=\frac{6^{\frac{1}{x}}}{6^{\frac{1}{2x}}+3}$$

Variable h cannot be equal to 0.

$$h=\frac{6^{\frac{1}{x}}}{6^{\frac{1}{2x}}+3}\text{, }h\neq 0$$

Solve for x

$$x=\frac{\ln(6)}{2\left(\ln(\sqrt{h^{2}+12h}+h)-\ln(2)\right)}$$

$$h\neq \frac{1}{4}\text{ and }h>0$$