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I am trying to convert $1.5\times10^{8}$ to a normalized floating point system defined as $0.d_1d_2d_3...d_t\times B^c$ for $t=20, B=2$

I don't know how I can change my base 10 to a base 2 without getting a non-integer value for $c$ (I am not sure if c has to be an integer value or not).

To get $t=20$, I could do something crazy like $0.15000000000000000000\times10^9$. But then again my problem arises from trying to convert my base. I do not know what the right way is to approach this problem.

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For the exponent $c$ that has indeed to be integer, you compute the next largest integer to the binary logarithm, $\log_2(1.5\cdot 10^8)=27.160..$ Then for the mantissa you need the integer part of $1.5\cdot 10^8\cdot 2^{20-28}=585937.5$. Now to get the digits you need to convert this decimal number to binary. One step there could be the hexadecimal form ${\tt 8F0D2}$.

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