Solve (3x^2-5)^4=ax^b

$$( 3 x ^ { 2 } - 5 ) ^ { 4 } = a x ^ { b }$$

Solve for a

$$a=\frac{\left(3x^{2}-5\right)^{4}}{x^{b}}$$

$$x>0\text{ or }\left(Denominator(b)\text{bmod}2=1\text{ and }x<0\right)$$

Steps for Solving Linear Equation

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

$$ax^{b}=\left(3x^{2}-5\right)^{4}$$

The equation is in standard form.

$$x^{b}a=\left(3x^{2}-5\right)^{4}$$

Divide both sides by x^{b}.

$$\frac{x^{b}a}{x^{b}}=\frac{\left(3x^{2}-5\right)^{4}}{x^{b}}$$

Dividing by x^{b} undoes the multiplication by x^{b}.

$$a=\frac{\left(3x^{2}-5\right)^{4}}{x^{b}}$$

Solve for b

$$\left\{\begin{matrix}b=\log_{x}\left(\frac{\left(3x^{2}-5\right)^{4}}{a}\right)\text{, }&a>0\text{ and }x\neq 1\text{ and }x>0\text{ and }x\neq \frac{\sqrt{15}}{3}\\b\in \mathrm{R}\text{, }&\left(a=0\text{ and }x=\frac{\sqrt{15}}{3}\right)\text{ or }\left(a=16\text{ and }x=1\right)\text{ or }\left(a=0\text{ and }Denominator(b)\text{bmod}2=1\text{ and }x=-\frac{\sqrt{15}}{3}\right)\text{ or }\left(a=-16\text{ and }Denominator(b)\text{bmod}2=1\text{ and }Numerator(b)\text{bmod}2=1\text{ and }x=-1\right)\end{matrix}\right.$$