Precalculus Exponential and Logarithmic Functions
Properties of Logarithms Concepts 1

|  |  | Statement 1 |  |
. |
Logarithm - of - a - Product Property log a (x·y) = loga x + loga y

Example 1. Using log103 = 0.4771 and log105 = 0.6990 , compute log1015 .

Solution

log 10 15 | = log10 (3·5)
| = log10 3 + log10 5
| = 0.4771 + 0.6990
| = 1.1761 .

Answer: log 10 15 = 1.1761 .

|  |  | Statement 2 |  |
. |
Logarithm - of - a - Quotient Property

Example 2. Using log103 = 0.4771 , compute .

Solution

| =
| = log10 10 - log10 3
| = 1 - 0.4771
| = 0.5229 .

Answer: = 0.5229 .

|  |  | Statement 3 |  |
. |
Logarithm - of - a - Power Property loga (xb) = b·loga x

Example 3. Using log102 = 0.3010 , compute log1016 .

Solution

log10 16 | = log10 (24)
| = 4·log102
| = 4·0.3010
| = 1.2040 .

Answer: log10 16= 1.2040 .

|  |  | Statement 4 |  |
. |
Change of Base Formula

Example 4. Using ln10 = 2.3026 and ln5 = 1.6094 , compute log105 .

Solution

log105 | =
| =
| = 0.6990 .

Answer: log 10 5 = 0.6990 .

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