how can i solve log 0.9. please give step by step solution?

Answer 1

log 0.9 is not anything that needs to be "solved". You just look it up on a table or use a calculator.

The easiest is to type Log (0.9) into the Google search engine and you get:

log(0.9) = -0.0457574906

Answer 2

Log (to the base 10) of [10] = 1.
This is because the laws of logarithms (special logarithms) say:
Log (to the base x ) of [x ] = 1.

Log (to the base 10) of [100] = 2.
This is because 100 = 10^2. 2 is called the exponent.

Log (base 10) of [0.9] = 10^0.9.
If you plug into Yahoo! Web Search,
10^(2) you get 100 out
10^(1) you get 10 out
10^(0.9) you get 7.94 out [ans].

Answer 3

Thanks for giving a very simple question to us. Follow my step below.

Firstly, given log 0.9.
We know that 0.9 = 9/10

So, log 0.9 = log (9/10)
log 0.9 = log 9 - log 10

Next, we solve it by using calculator:
log 9 = 0.95424
log 10 = 1

log 9 - log 10 = 0.95424 - 1
log 9 - log 10 = -0.04576

Therefore, the answer is -0.04576

Answer 4

9^21 = 109,418,989,131,512,359,209 > 10^20
by 9.42%
9^22 = 984,770,902,183,611,232,881 < 10^21
by 1.52%
Using 9^22 ≈ 10^21,
log(9^22) ≈ log(10^21)
22log(9) ≈ 21log(10)
22log(9) ≈ 21
log(9) ≈ 21/22
log(9) ≈ 0.9545455
log(0.9) = log(9/10) = log(9) - log(10) = log(9) - 1
log(0.9) ≈ - 0.04545455

This is within 0.67% of the value
- 0.045757490560675125409944193489769
arrived at by calculator.

Answer 5

Your question makes no sense. What do you mean by "solving" log 0.9? Perhaps you should brush up on your language usage.

Answer 6

Unless you want to calculate it from an infinite series you either use a calculator or log tables.

If you do want to calculate it from an infinite series, look up McLauren's theorem, and then work out the series for log[ (1-x)/(1 + x)].