Question

Solve for x with steps logb(8)+logb(x^2)=logb(x)

Answer 1

Answer:

$x=\frac{1}{8}$

Step-by-step explanation:

$\log _b\left(8\right)+\log _b\left(x^2\right)=\log _b\left(x\right)$

$\log _b\left(8\right)+2\log _b\left(x\right)=\log _b\left(x\right)$

$\log _b\left(8\right)+2\log _b\left(x\right)-\log _b\left(8\right)=\log _b\left(x\right)-\log _b\left(8\right)$

$2\log _b\left(x\right)=\log _b\left(x\right)-\log _b\left(8\right)$

$2\log _b\left(x\right)-\log _b\left(x\right)=\log _b\left(x\right)-\log _b\left(8\right)-\log _b\left(x\right)$

$\log _b\left(x\right)=-\log _b\left(8\right)$

$\log _b\left(x\right)=-3\log _b\left(2\right)$

$x=b^{-3\log _b\left(2\right)}$$x=b^{-3\log _b\left(2\right)}$

$b^{-3\log _b\left(2\right)}$

$=\left(b^{\log _b\left(2\right)}\right)^{-3}$

$=2^{-3}$

$=\frac{1}{8}$

$x=\frac{1}{8}$