Question

K digits are required in base b for a number ,how many digits are required in base x ?

Answer 1

Largest number in base $b = b^k-1$.

Let $l$ be the required no. of bits in base $x$ which gives the largest number as $x^l-1$

We need $x^l - 1 \geq b^k -1 \\ \implies x^l \geq b^k \\ \implies l \log x \geq k \log b \\ \implies l \geq \frac{k \log b}{\log x}$

Comments

Sir what is the reason behind this inequality which you have used , why can't be <= or >= ?

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Sorry. It must be >=. Why not <? because we want to ensure all numbers representable in base b using k bits must be represantable in base x- there is no problem in having more bits but minimum we need this many (>=) bits.

Answer 2

log b^k base x = k (log b )/log x