Question

How do you solve `Log_(5)x + log_(3)x = 1`?

Answer 1

`x=e^((ln5(ln3))/(ln3+ln5))approx1.9211`

Explanation:

Use the change of base formula, which states that

`log_ab=log_cb/log_ca`

The common base I'll use here is `e`, since it is the base of the natural logarithm. It doesn't actually matter which base is chosen: any base will work.

The original expression can be rewritten as:

`lnx/ln5+lnx/ln3=1`

Find a common denominator of `ln5(ln3)`:

`(lnx(ln3))/(ln5(ln3))+(lnx(ln5))/(ln5(ln3))=1`

`(lnx(ln3)+lnx(ln5))/(ln5(ln3))=1`

Cross multiply.

`lnx(ln3)+lnx(ln5)=ln5(ln3)`

Factor a `lnx` from both terms on the left hand side.

`lnx(ln3+ln5)=ln5(ln3)`

Divide both sides by `ln3+ln5`.

`lnx=(ln5(ln3))/(ln3+ln5)`

To undo the natural logarithm, exponentiate both sides with base `e`.

`x=e^((ln5(ln3))/(ln3+ln5))approx1.9211`