Question

Evaluate ㏒√35 +㏒√2 +㏒√7

Answer 1

Answer:

The answer is

$\frac{1}{2} + log(7)$

Step-by-step explanation:

$log( \sqrt{35} ) + log( \sqrt{2} ) + log( \sqrt{7} )$

We use the rules of logarithms.

Since they have the same base which is base 10 we multiply the numbers.

That's

$log(x) + log(y) = log(x \times y)$

So we have

$log( \sqrt{35} \times \sqrt{ 2} \times \sqrt{7} ) \\ = log( \sqrt{35 \times 2 \times 7} ) \\ = log( \sqrt{490} )$

Simplify the surd

Thats

$\sqrt{490} = 7 \sqrt{10}$

So we have

$log(7 \sqrt{10} )$

Next using the rules of logarithms expand the expression

That's

$log(x \times y) = log(x) + log(y)$

So we have

$log(7 \sqrt{10} ) = log(7) + log( \sqrt{10} )$

Write the expression in exponential form

That's

$\sqrt{10} = {10}^{ \frac{1}{2} }$

$log(7 \sqrt{10} ) = log(7) + log( {10})^{ \frac{1}{2} } \\ \\ = log(7) + \frac{1}{2} log(10)$

But

$log(10) = 1$

Simplify

We have the final answer as

$\frac{1}{2} + log(7)$

Hope this helps you.