Question

Solve. Log8 (6x) = Log8 2 + Log8 (x-4)

Answer 1

Write both sides as powers of 8, then simplify:

$8^{\log_86x}=8^{\log_82+\log_8(x-4)}=8^{\log_82}8^{\log_8(x-4)}$

$\implies6x=2(x-4)$

$\implies6x=2x-8$

$\implies4x=-8$

$\implies x=-2$

However, on the left hand side, this would mean we'd have, $\log_8-12$, which is undefined (assuming the real-valued logarithm), so this equation has no real solution.