Question

the logarithm of 21952 to the base of 2√7 and 19683 to the base of 3√3 are​

Answer 1

Answer:

The logarithm of 21952 to the base of 2√7 is 6 and the logarithm 19683 to the base of 3√3 is 6.

Step-by-step explanation:

Logarithm of 21952 to the base of 2√7:

Let’s break down 21952 in terms of 2√7

21952 = 2*2*2*2*2*2*√7*√7*√7*√7*√7*√7=(2√7)⁶

Now, we can write logarithm of 21952 to the base of 2√7 as

log(base 2√7){(2√7)⁶}……[∵ log (base b) a = log b / log a]

= {log(2√7)⁶} / { log 2√7 }

= 6 log(2√7) / log(2√7) ……[∵ log a^b = b log a]

= 6

Logarithm of 19683 to the base of 3√3:

Let’s break down 19683 in terms of 3√3

21952 = 3*3*3*3*3*3*√3*√3*√3*√3*√3*√3=(3√3)⁶

Now, we can write logarithm of 19683 to the base of 3√3 as

log(base 3√3){( 3√3)⁶}

= {log(3√3)⁶} / { log 3√3}

= 6 log(3√3) / log(3√3)

= 6

Answer 2

Solution:

As we know that from the property of Logarithmic

$log_{m}(m) = 1 \\ \\ log_{m}( {m}^{n} ) = n log_{m}(m) = n$
Here in the given question ,if somehow we can write the given function in terms of base than we can easily Solve this

$log_{2 \sqrt{7} }(21952) \\ \\ = > log_{2 \sqrt{7} }( {2}^{6} \times {7}^{3} ) \\ \\ = > log_{2 \sqrt{7} }( {2}^{6} \times ({ \sqrt{7} })^{6} ) \\ \\ = > log_{2 \sqrt{7} }( {(2 \sqrt{7} })^{6} ) \\ \\ = > 6log_{2 \sqrt{7} }(2 \sqrt{7} ) \\ \\ log_{2 \sqrt{7} }(21952)= > 6$
By the same way

$log_{3 \sqrt{3} }(19683) \\ \\ = > log_{3 \sqrt{3} }( {3}^{9} ) \\ \\ = > log_{3 \sqrt{3} }( {3}^{6} \times {3}^{3} ) \\ \\ = > log_{3 \sqrt{3} }( {3}^{6} \times { \sqrt{3} }^{6} ) \\ \\ = > log_{3 \sqrt{3} }( {3 \sqrt{3} })^{6} \\ \\ = > 6 \: log_{3 \sqrt{3} }( {3 \sqrt{3} }) \\ \\ log_{3 \sqrt{3} }(19683)= > 6$
Hope it helps you.