How to Solve Logarithmic Equations Without Extra Answers

Logarithms and exponents are inverses:

lo g_{b} (y) = x means b^{x} = y

You can only take the logarithm of a positive number. That's why logarithmic equations can produce answers that don't work.

Rewrite: $2 x + 1 = 3^{2} = 9$

$x = 4$

Check: $2 (4) + 1 = 9$ is positive, so it works.

Combine: $lo g_{2} (x (x - 2)) = 3$

Rewrite: $x (x - 2) = 8$

$x^{2} - 2 x - 8 = 0$ , so $x = 4$ or $x = - 2$

$x = - 2$ makes $lo g_{2} (x)$ undefined, so the only solution is $x = 4$ .

$lo g (x + 3) = lo g x + lo g 3$

The product rule says $lo g (ab) = lo g a + lo g b$ . There is no rule that splits the logarithm of a sum.

Try it yourself