# INTRODUCTION TO LOGARITHM (LAWS OF LOGARITHM)

Logarithms are the inverse function of exponentiation.

This implies the logarithm of a given number $y$ is the exponent to which another fixed number, $a$ must be raised, to produce that number $y$.

Now, let's look at the three basic laws of logarithmic .

### First law of logarithm

$\log _{b}s+\log_{b}t=\log_{b}st$

### Second law of logarithm

$\log_{b}s-\log_{b}t=\log_{b}\frac{s}{t}$

### Third law of logarithm

$\log_{b}s^t=t\log_{b}s$

Forthwith, let's try some examples.

#### Example 1

Solve $\log_{3}27$

Solution:

Let $\log_{3}27=x$

Index form

$27=3^x$

$3^3=3^x$

$3=x$

$x=3$

Hence

$\log_{3}27=3$

#### Example 2

$\log_{4}x=3$

Solution:

$\log_{4}x=3$ becomes $x=4^3$

So $x=64$

#### Example 3

Write $\log_{}800$ in terms of log 4, log 5, log 8 to any base.

Solution:

$\log_{}800=\log_{}(25\times32)=\log_{}(25\times 8\times4)$

$\log_{}(5^2\times8\times4)$

$\log_{}5^2+\log_{}8+\log_{}(4)$

By the first law of logarithms

$\log_{}800=2\log_{}5+\log_{}8+\log_{}4$

By the third law of logarithm

#### Example 4

Simplify  $\log_{2}1024+\log_{2}256-\log_{2}512$

Solution:

$1024=2^{10}$, $256=2^8$,  $512=2^9$

Hence,

$\log_{2}1024+\log_{2}256-\log_{2}512$

$=\log_{2}2^{10}+\log_{2}2^8-\log_{2}2^9$

$=10\log_{2}2+8\log_{2}2-9\log_{2}2$

By the third law of logarithm

If $\log_{2}2=1$, Then,

$10\log_{2}2+8\log_{2}2-9\log_{2}2$

$10(1)+8(1)-9(1)$

$10+8-9=9$

#### Example 5

Evaluate $\log_{6}216-\log_{¼}256+\log_{5}3125$

Solution:

The base is not the same, so we will simplify separately before evaluating

Let $\log_{6}216=a$

Index form

$216=6^a$

$6^3=6^a$

$3=a$

$a=3$

Let $\log_{¼}256=b$

Index form

$256=¼^b$

$4^4=4^{-1(b)}$

$4^4=4^{-b}$

$4=-b$

Divide both sides by $-1$

$\frac{4}{-1}=\frac{-b}{-1}$

$-4=b$

$b=-4$

Let $\log_{5}3125$=c

Index form

$3125=5^c$

$5^5=5^c$

$5=c$

$c=5$

Having gotten the value of all log, let's evaluate together

$\log_{6}216-\log_{¼}256+\log_{5}3125$

$=3-(-4)+5=12$

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#### Example 6

Evaluate

$\log_{7}x=\log_{16}256$

Solution:

$\log_{7}x=\log_{16}16^2$

$\log_{7}x=2\log_{16}16$

If $\log_{16}16=1$, then

$\log_{7}x=2(1)$

$\log_{7}x=2$

Index form

$x=7^2$

$x=49$

#### Example 7

Simplify

$\log_{10}(10+3\sqrt{10})+\log_{10}(10+(\sqrt{90+\sqrt{90}}+\log_{10}(10-(\sqrt{90+\sqrt{90}}))$

Solution:

To simplify this, we multiply the middle log to the right-hand log

$\log_{10}(10+3\sqrt{10})+\log_{10}(10+\sqrt{90+\sqrt{90}})(10-(\sqrt{90+\sqrt{90}}$)

By taking the difference of two squares, this will translates into

$\log_{10}(10+3\sqrt{10})+\log_{10}(100-({\sqrt{90+\sqrt{90}}}){^2}$

Note: Square cancels square root

$\log_{10}(10+3\sqrt{10})+\log_{10}(100-(90+\sqrt{90}$

Now, let's open the bracket

$\log_{10}(10+3\sqrt{10})+\log_{10}(100-90-\sqrt{90})$

$\log_{10}(10+3\sqrt{10})+\log_{10}(10-\sqrt{9\times10})$

$\log_{10}(10+3\sqrt{10})+\log_{10}(10-3\sqrt{10})$

$\log_{10}(10+3\sqrt{10})(10-3\sqrt{10})$

By taking the difference of two squares, this translates into

$\log_{10}(100-(9(\sqrt{10}^2)$

$\log_{10}(100-(9\times10)$

$\log_{10}(100-90)$

$\log_{10}10=1$

#### Example 8

Solve $3\log_69+\log_612+\log_664-\log672$

Solution:

$\log_69^3+\log_612+\log_664-log_672$

$\log_6729+\log_612+\log_664-log_672$

$\log_6\frac{729×12×64}{72}$

$\log_67776$

$\log_66^5$

$5\log_66=5(1)=5$

I just introduced logarithm with these 7 examples, In our next posts, you will see some examples that will solidify your knowledge of logarithm, be sure to check it out here.

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