Logarithms - Solved Examples

Solved Example 1: (Important model)
How many digits are contained in the number 2100$2^{100}$
Sol: log2100$\log 2^{100}$ = 100 x log 2 = 100 x 0.3010 = 30.10
Number of digits in 2100$2^{100}$ are 30 + 1 = 31

To determine the characteristic of the logarithm of a decimal fraction: (Numbers between 0 to 1)
Look at this example:
Find the total zeroes after he decimal point of the expression 26$2^6$
We know that 26$2^6$ = 164${\displaystyle \frac{1}{64}}$ = 0.015625
log 164${\displaystyle \frac{1}{64}}$  = -1.806
Now when you calculate Antilog of -1.806 using calculator, you will get 0.0156.
But if you want to use antilog tables, you have to follow this procedure.
Now log 164${\displaystyle \frac{1}{64}}$ = log 126${\displaystyle \frac{1}{2^6}}$ = log 2−6$2^{-6}$ = -6 log 2.
We know that log 2 = 0.301
Now log 164${\displaystyle \frac{1}{64}}$ = -6 × 0.301 = -1.806.
Important: Now if you look at the antilog table for 0.80 and 6, you will get wrong answer.  Why? Because -1.806 = -1 + (-0.806)
But mantissa is always positive.
-1.806 should be written as -2 + (1 - 0.806) = -2 + 0.194
Now when you look at the anti log table 0.194 gives you 1563. 
So characteristic is 2.

That is the characteristic of the logarithm of a decimal fraction is one more than than the number of zeroes immediately after the decimal point  and is negative.

Working Rule to find the number of zeroes in a decimal number: 
1. Calculate the logarithm (you will get some negative number)
2. Subtract the decimal part from one and increase the integer part by 1 to make mantissa positive
3. Number of zeroes of that number = Integer part - 1

Solved Example 2: (Important model)
How many zeroes are there between the decimal point and the first significant digit in (12)1000${\left({\displaystyle \frac{1}{2}}\right)}^{1000}$
log (12)1000${\left({\displaystyle \frac{1}{2}}\right)}^{1000}$ = 1000 × log (1/2) = 1000 × -0.30102 = -301.02
But in logarithms the decimal point should be positive. (By using
- 301.02 = - 301 + -0.02 = -302 + (1 - 0.02) = 302___$\stackrel{\mathrm{\_}\mathrm{\_}\mathrm{\_}}{302}$.98
So number of zeroes are 302 - 1 = 301

Solved Example 3:
11+logabc+11+logbca+11+logcab=${\displaystyle \frac{1}{1+{\log }_{a}bc}+{\displaystyle \frac{1}{1+{\log }_{b}ca}+{\displaystyle \frac{1}{1+{\log }_{c}ab}=}}}$

a. 0	b. 3
c. 2	d. 1

Answer: d
Explanation:
11+logabc${\displaystyle \frac{1}{1+{\log }_{a}bc}}$ + 11+logbca${\displaystyle \frac{1}{1+{\log }_{b}ca}}$ + 11+logcab${\displaystyle \frac{1}{1+{\log }_{c}ab}}$= 
1logaa+logabc${\displaystyle \frac{1}{{\log }_{a}a+{\log }_{a}bc}}$+ 1logbb+logbca${\displaystyle \frac{1}{{\log }_{b}b+{\log }_{b}ca}}$ + 1logcc+logcab${\displaystyle \frac{1}{{\log }_{c}c+{\log }_{c}ab}}$
= 1logaabc${\displaystyle \frac{1}{{\log }_{a}abc}}$ + 1logbabc${\displaystyle \frac{1}{{\log }_{b}abc}}$ + 1logcabc${\displaystyle \frac{1}{{\log }_{c}abc}}$ = 
= logabca+logabcb+logabcc=logabcabc=1${\log }_{abc}a+{\log }_{abc}b+{\log }_{abc}c={\log }_{abc}abc=1$

Solved Example 4:
The value of  (yz)logy−logz×(zx)logy−logx×(xy)logx−logy$(yz{)}^{\log y-\log z}\times (zx{)}^{\log y-\log x}\times (xy{)}^{\log x-\log y}$

a. 2	b. 1
c. 0	d. 3

Answer: b
Explanation:
Assume K = (yz)logy−logz$(yz{)}^{\log y-\log z}$ × (zx)logy−logx$(zx{)}^{\log y-\log x}$ × (xy)logx−logy$(xy{)}^{\log x-\log y}$
Taking log on both sides
Log K = log ((yz)logy−logz$(yz{)}^{\log y-\log z}$ × (zx)logy−logx$(zx{)}^{\log y-\log x}$ × (xy)logx−logy$(xy{)}^{\log x-\log y}$)
= log(yz)logy−logz$\log (yz{)}^{\log y-\log z}$ + log(zx)logy−logx$\log (zx{)}^{\log y-\log x}$ + log(xy)logx−logy$\log (xy{)}^{\log x-\log y}$
= (logy−logz)log(yz)$(\log y-\log z)\log (yz)$ + (logz−logx)log(zx)$(\log z-\log x)\log (zx)$ + (logx−logy)log(xy)$(\log x-\log y)\log (xy)$
= (logy−logz)(logy+logz)$(\log y-\log z)(\log y+\log z)$ + (logz−logx)(logz+logx)$(\log z-\log x)(\log z+\log x)$ + (logx−logy)(logx+logy)$(\log x-\log y)(\log x+\log y)$ =0
log K = 0⇒$\Rightarrow$ K = 1

Solved Example 5:
log(x+y3)$\log \left({\displaystyle \frac{x+y}{3}}\right)$ = 12${\displaystyle \frac{1}{2}}$ (logx + logy)  then (xy+yx${\displaystyle \frac{x}{y}}+{\displaystyle \frac{y}{x}}$) is

a. 5	b. 7
c. 9	d. 0

Answer: b
Explanation: 
As the answer does not have any log, first we try to remove log from the given equation by simplifying it.
log(x+y3)$\left({\displaystyle {\displaystyle \frac{x+y}{3}}}\right)$ =  12${\displaystyle \frac{1}{2}}$(logx + logy)
⇒2log(x+y3)$\Rightarrow 2log\left({\displaystyle \frac{x+y}{3}}\right)$= log xy
⇒$\Rightarrow$ log(x+y3)2=logxy$\log {\left({\displaystyle \frac{x+y}{3}}\right)}^2=\log xy$
⇒$\Rightarrow$x2+y2+2xy=9xy$x^2+y^2+2xy=9xy$
⇒$\Rightarrow$x2+y2=7xy$x^2+y^2=7xy$
⇒$\Rightarrow$ xy+yx=7