Question

The simplified value of log4√{729 3√(9-1.27-4/3)}is



a) log3

b)log2

c)log1/2

d) none of these

Answer 1

HOPEFULLY THE ATTACHMENT WILL HELP YOU.......

Answer 2

Option 1 - $\log[729(9^{-1}27^{-\frac{4}{3}})^{\frac{1}{3}}}}]^{\frac{1}{4}}=\log 3$

Step-by-step explanation:

Given : Expression $\log[\sqrt{729(9^{-1}27^{-\frac{4}{3}})^{\frac{1}{3}}}}]^{\frac{1}{4}$

To find : Simplify the expression ?

Solution :

Expression $\log[729(9^{-1}27^{-\frac{4}{3}})^{\frac{1}{3}}}}]^{\frac{1}{4}$

We simplify the bracket term first,

$(9^{-1}27^{-\frac{4}{3}})^{\frac{1}{3}}}}=(3^{-2}3^{-\frac{4\times 3}{3}})^{\frac{1}{3}}}}$

$(9^{-1}27^{-\frac{4}{3}})^{\frac{1}{3}}}}=(3^{-2}3^{-4})^{\frac{1}{3}}$

$(9^{-1}27^{-\frac{4}{3}})^{\frac{1}{3}}}}=(3^{-6})^{\frac{1}{3}}$

$(9^{-1}27^{-\frac{4}{3}})^{\frac{1}{3}}}}=3^{-2}$

$(9^{-1}27^{-\frac{4}{3}})^{\frac{1}{3}}}}=\frac{1}{9}$

Substitute back,

$\log[729(9^{-1}27^{-\frac{4}{3}})^{\frac{1}{3}}}}]^{\frac{1}{4}}=\log(729\times \frac{1}{9})^{\frac{1}{4}$

$\log[729(9^{-1}27^{-\frac{4}{3}})^{\frac{1}{3}}}}]^{\frac{1}{4}}=\log[81]^{\frac{1}{4}$

$\log[729(9^{-1}27^{-\frac{4}{3}})^{\frac{1}{3}}}}]^{\frac{1}{4}}=\log[3^4]^{\frac{1}{4}$

$\log[729(9^{-1}27^{-\frac{4}{3}})^{\frac{1}{3}}}}]^{\frac{1}{4}}=\log 3$

Therefore, option a is correct.

#Learn more

Simplify: 13375 x (-729).​

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