How do you evaluate log98.2?

3 Answers
Oct 11, 2017

Answer:

See explanation

Explanation:

These days people use calculators. Years ago log tables were used and I am not sure if I can even find my old copy of one. If you wish to use log tables I did a quick search and found this site.

https://www.wikihow.com/Use-Logarithmic-Tables

The 9 in 98.2 is counting in tens so you will have a log value starting as #1.0# plus some decimal that you put after the decimal point.

My calculator gives: 1.9921 rounded to 4 decimal places.
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What is #log98.2# actually saying?

When you see it written like this it is generally accepted that you are using what is called base10. Really it should be written as #log_10 98.2#

Suppose we set #log98.2=x#

Then this is stating the condition that #10^x=98.2#

EVALUATE means give value to. The value of log98.2 is

1.9921 rounded to 4 decimal places.

Oct 12, 2017

Answer:

#log(98.2) = log(100)-ln(1-0.018)/ln(10) ~~ 1.99211#

Explanation:

I will calculate this to just a few significant digits, but the same method can give you more using more terms...

Use:

#ln 10 ~~ 2.302585093#

#ln(1-t) = -t-t^2/2-t^3/3-t^4/4-...#

So:

#log(98.2) = log(100*0.982)#

#color(white)(log(98.2)) = log(100)+log(1-0.018)#

#color(white)(log(98.2)) = 2+ln(1-0.018)/ln(10)#

Now:

#ln(1-0.018) = -0.018-0.018^2/2-0.018^3/3-0.018^4/4-...#

#color(white)(ln(1-0.018)) ~~ -0.018-0.000324/2#

#color(white)(ln(1-0.018)) ~~ -0.018162#

So:

#log(98.2) ~~ 2-0.018162/2.3026 ~~ 1.99211#

Oct 12, 2017

Answer:

Use #log(2) ~~ 0.30103# to find #log(98.2) ~~ 1.992#

Explanation:

Use:

#log(2) ~~ 0.30103#

Then:

#98.2 = 100*0.982 = 100*(1-0.018) ~~ 100/(1+0.018) ~~ 100/(1.024)^(3/4) = 100/((2^10/10^3)^(3/4))#

So:

#log(98.2) ~~ log(100)-log(((2^10)/(10^3))^(3/4))#

#color(white)(log(98.2)) ~~ 2-3/4 (10log(2)-3)#

#color(white)(log(98.2)) ~~ 2-3/4 (3.0103-3)#

#color(white)(log(98.2)) ~~ 2-3/4 (0.0103)#

#color(white)(log(98.2)) ~~ 1.992#