Assuming 0x6A is given in base 16, first convert [tex]152_b[/tex] and [tex]6A_{16}[/tex] to a common base, say base 10:
[tex]152_b=1\cdot b^2+5\cdot b^1+2\cdot b^0=(b^2+5b+2)_{10}[/tex]
[tex]6A_{16}=6\cdot16^1+10\cdot16^0=106_{10}[/tex]
Then
[tex]b^2+5b+2=106[/tex]
[tex]b^2+5b-104=(b-13)(b+8)=0[/tex]
[tex]\implies b=13[/tex]
Answer:
The value of b is, 8
Step-by-step explanation:
Determine the value of b;
Given:
[tex](152)_b = 0x6A[/tex] ....[1]
Since, 0x6A represents the hexadecimal form.
First convert this hexadecimal form into decimal form:
[tex](6A) = (6 \times 16^1)+(A \times 16^0) = (96)+(10 \times 1) = 96+10 = 106[/tex]
⇒[tex]0x6A = 106[/tex] (decimal form)
Now we have to convert this decimal form into octal
8 | 106
8 | 13 | 2
| 1 | 5
1
Then, the octal form we get, [tex](152)_8[/tex]
Substitute these in [1] we have;
[tex](152)_b=(152)_8[/tex]
On comparing both sides we get;
b = 8
Therefore, the value of b is, 8
