SOLUTION: Dear Sir/Madam,
Please help me with this logarithm problem.
If log2a + logbb ≥ 6, then the least value of a + b is.
note: the bases are 2 and b
Thank you very
Algebra.Com
Question 894861: Dear Sir/Madam,
Please help me with this logarithm problem.
If log2a + logbb ≥ 6, then the least value of a + b is.
note: the bases are 2 and b
Thank you very much!
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
logbb can be shown as log(b,b)
log2a can be shown as log(2,a)
your equation is log(2,a) + log(b,b) >= 6
since log(b,b) will always be equal to 1, then b can be any legitimate value greater than 0.
since log(b,b) will always be equal to 1, your equation becomes:
log(2,a) + 1 >= 6
subtract 1 from both sides of this equation to get log(2,a) >= 5
solve for log(2,a) = 5 to get log(2,a) = 5 if and only if 2^5 = a which occurs when a = 32.
so log(2,32) = 5
that appears to be the minimum value for a.
if a > 32, then log(2,a) will be greater than 5
if a < 32, then log(2,a) will be less than 5.
for example:
log(2,64) = 6 which is greater than 5.
log(2,16) = 4 which is less than 5.
a has to be greater than or equal to 32 and b has to be greater than or equal to 0.
therefore, a + b has to be greater than or equal to 32.
that's your solution as far as i can tell.
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