Question

2log5+log36-log9 pls euavalute pls I show work so I can learn please

Answer 1

$2\log5+\log36-\log9\\\\\text{use}\ n\log_ab=\log_ab^n\\\\=\log5^2+\log36-\log9\\\\\text{use}\ \log_ab+\log_ac=\log_(bc)\ \text{and}\ \log_ab-\log_ac=\log_a\dfrac{b}{c}\\\\=\log(25\cdot36)-\log9=\log\dfrac{25\cdot36}{9}\\\\\log\dfrac{25\cdot4}{1}=\log100=2\\\\\text{because}\ 10^2=100$

Answer 2

Answer:

2

Step-by-step explanation:

Actually, your job here is to simplify the given expression by consolidating three logs into one logarithmic expression.

Note that 2 log 5 = log 5^2 = log 25

Then we have log 25 + log 36 - log 9, which is equivalent to log (25)(36)/9, or log 25(4), or log 100, or log 10^2, or just 2.