**4.Use Log Rules to simplify the Following questions: **

**1. log _{a} a^{3} 2. log _{3} ^{3}√(81) **

*Solution 1:*

Let us use the third Rule of Logarithm given above:

log _{a} (p)^{n} = n log _{a} p

log _{a} a^{3}
= 3 log _{a} a

= 3 × 1 = 3 ( because log _{a} a = 1)

*Solution 2:*

First of all,

3√(81) = 3√(3^{4}) = (3^{4})^{1/3} = 3^{4/3}

So, log _{3} ^{3}√(81) = log _{3} 3^{4/3}

= (^{4}/_{3}) × log _{3} 3 = (^{4}/_{3}) ×1 (because log _{3} 3 = 1)

= ^{4} / _{3}.

**4. Find the values of the following **

**1. 3 ^{3 – log 37 } 2. [ 8^{log 2 3√11}]^{1/3}**

*Solution1:*

First recall laws of exponents:

a^{m – n} = a^{m} / _{an}

So, 3^{3 – log 37}

= (3)^{3} / (3)^{log 37}

= (27) / (3)^{log 37} ……..(1)

Now, apply the formula: a^{log a p} = p in the denominator for 3 ^{log 37}

So, 3^{log 37} = 7

Put this value 7 in the denominator in (1), we get

3^{3 – log 37} = ^{27}/_{7}

**2. [8 ^{log 2 3√11}]^{1/3}**

*Solution2:*

First of all,

8^{log 2 3√11}

= (8)^{ log 2 3√11} { because a^{m} is same as (a)^{m} }

= (2^{3})^{(log 2 3√11)}

= (2)^{3log 2 3√11} {recall that (a^{m})^{n} = a^{mn} }

Now [8 ^{log 2 3√11]1/3}

= [(2)^{3log 2 3√11)}]^{1/3}

= [2^{3log 2 3√11)}]^{1/3}

Recall laws of exponents again,

(a^{m})^{ 1/n} = a^{m/n}

So, [2^{3 log 2 3√11)}]^{ 1/3}

= [2^{3(log 2 3√11)/3}]

= 2^{(log 2 3√11)} {because 3^{(log 2 3√11)/3} = log _{2} 3√11}

Now, apply apply the formula

a^{(log a p)} = p on 2^{(log 2 3√11)}

We get 2^{(log 2 3√11)} = 3√11

Therefore,

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