Question

If given
`\log_p(3) \approx 0.14` and
`\log_p(2) \approx 0.09` and
`log_p(5) \approx 0.21`, then find
`\log_p(60)` approximately.

Answer 1

Heya!

We can use the extended version product rule to solve: log_p(abc) = log_p(a) + log_p(b) + log_p(c)

We can factor 60 using 2, 3, and 5.

60 = 2^2 * 3 * 5

Substitute these numbers into the rule.

log(2^2 * 3 * 5) = 2 log_p(2) + log_p(3) + log_p(5)

Using the values given in the question, we can substitute in the equation.

2 log_p(2) + log_p(3) + log_p(5) → 2(0.09) + 0.14 + 0.21

Simplify the equation.

2(0.09) + 0.14 + 0.21

0.18 + 0.14 + 0.21

0.53

Therefore,
`\text{log}_p(60) = 0.53`

Best of Luck!