Solve 2^x+4=50^x+1quadh)

$$2 ^ { x + 4 } = 50 ^ { x + 1 } \quad h )$$

Solve for h (complex solution)

$$h=\frac{8\times \left(\frac{1}{25}\right)^{x}}{25}$$

Steps for Solving Linear Equation

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

$$50^{x+1}h=2^{x+4}$$

Divide both sides by 50^{x+1}.

$$\frac{50^{x+1}h}{50^{x+1}}=\frac{2^{x+4}}{50^{x+1}}$$

Dividing by 50^{x+1} undoes the multiplication by 50^{x+1}.

$$h=\frac{2^{x+4}}{50^{x+1}}$$

Divide 2^{4+x} by 50^{x+1}.

$$h=\frac{8\times 2^{x}}{25\times 50^{x}}$$

Solve for h

$$h=\frac{8}{25\times 5^{2x}}$$

Steps for Solving Linear Equation

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

$$50^{x+1}h=2^{x+4}$$

Divide both sides by 50^{x+1}.

$$\frac{50^{x+1}h}{50^{x+1}}=\frac{2^{x+4}}{50^{x+1}}$$

Dividing by 50^{x+1} undoes the multiplication by 50^{x+1}.

$$h=\frac{2^{x+4}}{50^{x+1}}$$

Divide 2^{4+x} by 50^{x+1}.

$$h=\frac{8\times 2^{x}}{25\times 50^{x}}$$

Solve for x

$$x=\frac{-\ln(h)+\ln(\frac{8}{25})}{2\ln(5)}$$

$$h>0$$