How to find the radix (base) of a number given its representation in another radix (base)?

By definition $$144_8=1\cdot8^2+4\cdot8^1+4\cdot8^0=64+32+4=100\;,$$ and $$121_r=1\cdot r^2+2\cdot r^1+1\cdot r^0=r^2+2r+1\;.$$ To find $r$, just solve the quadratic equation $$r^2+2r+1=100$$ by whatever method you find most convenient; the slickest method is probably to notice that $r^2+2r+1=(r+1)^2$, so $(r+1)^2=100=10^2$.

remember the bold letters are the powers

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(121)r=(144)8

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Now r and 8 is the base so,

$1\cdot r^2+2 \cdot r^1+1\cdot r^0 = 1 \cdot 8^2+4\cdot 8^1+4 \cdot 8^0$

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$r^2+2r+1=64+32+4$

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$r^2+2r+1=100$

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$(r+1)² =100$

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$r+1=10$

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$r= 10-1$

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$r=9$

First let's find what 144(8) equals, with "8" as the base Listing the digits in order, count them off from the right, starting with zero:

digits: 1 4 4 numbering: 2 1 0

Then the bottom row of numbers become the assigned exponential numbers, and so we have: 144= 1⋅8ˆ2 + 4⋅8ˆ1 + 4⋅8ˆ0 = 64 + 32 + 4 = 100

As for 121(r), with "r" as the base, we can again list off the digits in order, counting them off from the right starting with zero:

digits: 1 2 1 numbering: 2 1 0

Again, the bottom row of numbers become the assigned exponential numbers, and so we have: 121r= 1⋅rˆ2 + 2⋅rˆ1 + 1⋅rˆ0 = rˆ2 + 2r + 1

We can then find "r" by solving the quadratic equation that we have now created: rˆ2 + 2r + 1 = 100 equation (r + 1)ˆ2 = 100 = 10ˆ2 factor (r + 1) ± r + 1 = ± 10 use the square root property ± r = ± 10 - 1 isolate the "r" r = 9 solve for r

HOPE THIS HELPS!!! JUST CLARIFICATION OF ALL THE STEPS SO IT APPEARS LONGER THAN APPRECIATED :D