Question

If log3 base a= 2, why 'a' is an irrational? Give reason

Answer 1

Solution :-

$\sf log_a3 = 2$

Writing it in exponential form

$\implies \sf {a}^{2} = 3$

[ Because, if log N to the base a = x then a^x = N ]

Taking square root on both sides

$\implies \sf a = \pm\sqrt{3}$

But a ≠ - √3 since bases of logarithm cannot be negative

$\implies \sf a = \sqrt{3}$

We know that square root of any non perfect square is always irrational

3 is a non perfect square. So √3 is irrational

Therefore a = √3 is irrational.

Answer 2

Step-by-step explanation:

Given:

$log_{a}(3) = 2$

Further Simplifying, we get

$= > {a}^{2} = 3 \\ \\ = > a = \pm \sqrt{3}$

But, 'a' can't ve negative because it is a logarithmic expression.

Therefore, $\sf{a=\sqrt{3}}$

Here, The square of 'a' is a natural number which is not a perfect square.

Clearly, it will be the square of an irrational number.

Hence, 'a' is an irrational number.

Concept Map:-

• $\sf{log_{a}(m) = b \: \: \: \: = > {a}^{b} = m}$