what is the solution to $log_2(2x^{3}-8)-2log_2x = log_2x$?\n$x = 1$\n$x = 2$\n$x = 3$\n$x = 4$

Explanation:

Step1: Use log - rule $n\log_aM=\log_aM^n$

$\log_2(2x^3 - 8)-\log_2x^2=\log_2x$

Step2: Use log - rule $\log_aM-\log_aN=\log_a\frac{M}{N}$

$\log_2\frac{2x^3 - 8}{x^2}=\log_2x$

Step3: Since $\log_aM=\log_aN$ implies $M = N$

$\frac{2x^3 - 8}{x^2}=x$

Step4: Multiply both sides by $x^2$

$2x^3-8=x^3$

Step5: Rearrange the equation

$2x^3 - x^3=8$ $x^3=8$

Step6: Solve for $x$

$x = 2$

Answer:

$x = 2$