If log_3(a)=10 and log_3(b)=15, then log_3(a+b) is between ? and ?
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- Опубликовано: 30 май 2024
- This tutorial covers a logarithm problem that you might encounter in precalculus or algebra 2 classes. The question is if log_3(a)=10 and log_3(b)=15, then log_3(a+b) is between which two numbers? I posted this question on my Instagram, see the post here pC4Ros1bxY...
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This is a great problem, because the answer wasn't immediately obvious to me, but once I worked it out (essentially the same way as in the video), I saw why it should have been immediately obvious. It's similar to how, if you add a 10-digit number and a 15-digit number, the result is a 15-digit number (only in base 3 instead of base 10).
We can just directly apply inequality to solve it:
a + b = 3¹⁰ + 3¹⁵
< 3¹⁵ + 3¹⁵
< 3(3¹⁵)
= 3¹⁶
Hence log3(a + b) < 16, which gives answer A
And I guess you easily know it's at least 15 since one of the parts is already 15 even if an answer had a range below 15.
1:42 I factored out 3¹⁵, so was left with (1+3⁻⁵), which is just a little bit over 1. Its logarithm will be a little bit over 0.
That's way better than the approach in the video
I replaced the a+b with b+b=2b since b>a it would give us an upper bound and then applying the log laws we get log3(2b)=log3(2)+log3(b)
log3(2) is less than 1 and we know log3(b)=15 so the whole thing is less than 16
I knew by direct comparison that 10 < 15 means it's not sufficient to raise 3^15 up to 3^16, but it always helps to do more working out and checking the answer thoroughly, like BPRP! Logarithms are always a bit tricky, so doing more working out is always preferable to just winging it and assuming you're right.
Intuitively: it would take 2 more 3^15s to jump from 15th power to 16th power, 3^10 is much smaller than that so not enough to reach 16, therefore A must be the answer.
That last argument was a witness argument for Big-O.
This guy's a boss!
Solid
Not seen the vid yet, but for a log argument to increase such that the outcome is one higher, means to multiply with the base.
IOW: if log₃(27)=3, then log₃(3x27)=4. So, if log₃b=15, to increase to 16, you need log₃(3b). Since a must be lower than b, this is never true. Result is between 15 and 16 (and quite close to 15).
I think bprp took the long road home. The calculations weren't necessary here. But of course, result is the same.
I was like "wait a=3^10 and b=3^15 a+b must not exceed 3^16 does this ever need bprp to look into?"
Later "BASICS"