Question

Evaluate 3log2 + log200-log16

Answer 1

Answer:

2

Step-by-step explanation:

using the rules of logarithms

logx + logy = log(xy)

logx - logy = log ($\frac{x}{y}$ )

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 ($\frac{1600}{16}$ )

= log100

let $log_{10}$ 100 = n , then

100 = $10^{n}$ = 10²

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

Answer 2

$2 \cdot 1$Answer:

Equal to 2

Step-by-step 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: $log_c(a) - log_c(b) = log_c(\frac{a}{b})$

$log_{10}(1600) - log_{10}(16) = log_{10}(\frac{1600}{16})$

Divide the numbers: $\frac{1600}{16} = 100$

$= log_{10}(100)$

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: $log_a(a) = 1$

$log_{10}(10) = 1$

$= 2 \cdot 1 = 2$