Question

Evaluate 3log2 + log200-log16

Answer 1

`2 \cdot 1`Answer:

Equal to 2

Explanation:

1. Apply log rule:
`alog_c(b) = log_(c)(b^a)`

`3log_(10)(2) = log_(10)(2^3)`

`= log_(10)(2^3) + log_(10)(200) - log_(10)(16)`

2. Apply log rule:
`log_c(a) + log_c(b) = log_c(ab)`

`log_(10)(2^3) + log_(10)(200) = log_(10)(2^3 \cdot 200)`

`= log_(10)(2^3 \cdot 200) - log_(10)(16)`

`2^3 \cdot 200 = 1600`

`= log_(10)(1600) - log_(10)(16)`

3. Apply log rule:
`log_c(a) - log_c(b) = log_c((a)/(b))`

`log_(10)(1600) - log_(10)(16) = log_(10)((1600)/(16))`

Divide the numbers:
`(1600)/(16) = 100`

`= log_(10)(100)`

Rewrite 100 in power - base form:
`100 = 10^2`

`= log_(10)(10^2)`

4. Apply log rule:
`log_a(x^b) = b \cdot log_a(x)`

`log_(10)(10^2) = 2log_(10)(10)`

`2log_(10)(10)`

5. Apply log rule:
`log_a(a) = 1`

`log_(10)(10) = 1`

`= 2 \cdot 1 = 2`

Answer 2

2

Explanation:

using the rules of logarithms

logx + logy = log(xy)

logx - logy = log (
`(x)/(y)` )

log
`x^(n)` ⇔ nlogx

`log_(b)` x = n ⇒ x =
`b^(n)`

given

3log2 + log200 - log16

= log 2³ + log200 - log16

= log8 + log200 - log16

= log(8 × 200) - log16

= log1600 - log16

= log (
`(1600)/(16)` )

= log100

let
`log_(10)` 100 = n , then

100 =
`10^(n)` = 10²

thus value of expression is n = 2 , that is 2