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.