Question

find the value of
n= log 0.2 with base 10/log 0.12 with base 10

Answer 1

rule, log(m/n) = log(m) - log(n)

Method (1)
So, log(0.2) = log(2/10) = log(2) - log(10)
As, log(2) base_10 = 0.3010, and log(10) base_10 = 1
So, log(0.2) = 0.3010 - 1 = -0.6990

Method (2)
So, log(0.2) = log(2/10) = log(1/5) = log(1) - log(5)
As, log(1) base_10 = 0, and log(5) base_10 = 0.6990
So, log(0.2) = 0 - 0.6990 = -0.6990


rule, log(m * n) = log(m) + log(n)
rule, log(m * n * p) = log(m) + log(n) + log(p)

So, log(0.12) = log(12/100) = log(12) - log(100)
Now, log(12) = log(2 * 2 * 3) = log(2) + log(2) + log(3) = 2 * log(2) + log(3)
And log(100) = log(10 * 10) = log(10) + log(10) = 2 * log(10)
As, log(2) base_10 = 0.3010, log(3) base_10 = 0.4771, log(10) base_10 = 1
So, log(12) = 2 * 0.3010 + 0.4771 = 0.6020 + 0.4771 = 1.0791
And, log(100) = 2 * 1 = 2
So, log(0.12) = 1.0791 - 2 = -0.9209

Thus, log(0.2) / log(0.12) = -0.6990 / -0.9209 = 0.7590


Answer
0.7590

Answer 2

n = log 0.2  /  log 0.12        logarithms are with respect to base 10

log 0.2 = log 2 / 10  = log 2 - log 10  = log 2 - 1
log 0.12  = log 12/100 = log 3 * 2² /100 = log 3 + log 2² - log 100
               =  log 3 + 2 log 2 - 2

n = log 0.2 / log 0.12 
   =  [ log 2 - 1 ] / [ log 3 + 2 log 2 - 2 ]