Logarithm

The logarithm of a number is the power to which a given base must be raised to produce that number. For a given positive number a and a positive base b (where b ≠ 1), the logarithm of a with base b is written as $\log_{b} a$ (read as: log of a to the base b) and is defined as:

$\log_{b} a = c$ if and only if $b^{c} = a$

There are mainly two types of logarithm. They are:

Common logarithm
Natural logarithm

Common Logarithm: In 1624, an English mathematician Henry Briggs proposed a log table considering 10 as the base. This 10 based logarithm is known as common logarithm. It is also called Briggsian logarithm. The common logarithm became popular largely because of its alignment with the decimal number system. This similarity made it more practical for calculations in various fields.
Natural Logarithm: Natural logarithm is the logarithm where an irrational number e is considered as the base. The development of natural logarithm links back to John Napier. But the natural logarithm as we know it today was first explicitly connected to the base e by Nicholas Mercator.

Simplify

$7 \log_{10} \frac{10}{9} - 2 \log_{10} \frac{25}{24} + 3 \log_{10} \frac{81}{80}$
$\log_{7} (\sqrt[5]{7} \cdot \sqrt{7}) - \log_{3} \sqrt[3]{3} + \log_{4} 2$
$\log_{e} \frac{a^{3} b^{3}}{c^{3}} + \log_{e} \frac{b^{3} c^{3}}{d^{3}} + \log_{e} \frac{c^{3} d^{3}}{a^{3}} - 3 \log_{e} b^{2} c$
$\frac{\log_{10} \sqrt{27} + \log_{10} 8 - \log_{10} \sqrt{1000}}{\log_{10} 1.2}$ Ans. $\frac{3}{2}$
$w log \frac{xz}{y^{2}} - x log \frac{z^{2}}{x^{2} y} + y log \frac{y^{4}}{x^{4} z}$

You all know the name of corona virus. This virus spreads quickly. If corona virus spreads from 1 person to 3 persons per day, how many people will be infected with corona virus in 1 month? After how many days, 1 crore People will be infected?

Day	No. of person spreading the virus	Newly infected person	Total infected person
1	1	3*1=3	4=4^1
2	4	3*4=12	16=4^2
3	16	3*16=48	64=4^3
...	...	...	...
...	...	...	...
30			4^30

From the above calculations, we come to a conclusion that
Infected person in n-days = 4^n
∴ 1,00,000,00 = 4^n
apply logarithm formula.

Setu’s uncle has 3 bighas of land. He applies 30 kg of organic fertilizer every year to maintain the fertility of his land. If each kg of fertilizer increases the fertility of one katha land by 3%, find the depreciation of the land of Setu’s uncle? If he does not apply fertilizer to the land, then after how many years will his land have no crops?

Solution: We know,

P_{T} = {(1 - R)}^{T}

1 bigha = 20 katha > 3 bigha = 60 katha
1 kg fertilizer increase fertility of 1 katha land by 3 percent
1/2 kg fertilizer increase fertility of 1 katha land by 3*(1/2) = 1.5 percent
Assuming the fertility increase is to maintain current fertility levels, the annual natural depreciation (loss) in fertility must equal the fertility gained from the fertilizer.
∴ Rate of fertility decline or Depreciation of the land, R = 1.5% or, R = 0.015
Now,

0 = {(1 - R)}^{T}

Since we cannot derive a real value of T for the given condition, we can come to the conclusion that the land will never become cropless.

\frac{\log_{k} a}{b - c} = \frac{\log_{k} b}{c - a} = \frac{\log_{k} c}{a - b}

show that

a^{a} b^{b} c^{c} = 1

\frac{\log_{k} x}{p^{2} + pq + q^{2}} = \frac{\log_{k} y}{q^{2} + qr + r^{2}} = \frac{\log_{k} z}{r^{2} + rp + p^{2}}

show that

x^{p - q} \cdot y^{q - r} \cdot z^{r - p} = 1