SOLUTION: log[64](32)=0 I keep coming up with no solution, but that isn't one of the answers on my pretest. Thanks!

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Question 378530: log[64](32)=0
I keep coming up with no solution, but that isn't one of the answers on my pretest. Thanks!
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
Where's the variable?? Without a variable there is no "solution" to be found. The equation is either true or not.

As for whether
log%2864%2C+32%29+=+0
is true or not, just rewrite it in exponential form:
32+=+64%5E0
Since any non-zero number to the zero power is 1 this simplifies to:
32+=+1
This is clearly a false statement.

P.S. We can find what log%2864%2C+%2832%29%29 equals without using calculators. Since both 64 and 32 are powers of 2 (64+=+2%5E6 and 32+=+2%5E5, log%2864%2C+%2832%29%29 will simplify easily if we use the base conversion formula, log%28a%2C+%28p%29%29+=+log%28b%2C+%28p%29%29%2Flog%28b%2C+%28a%29%29, to convert this base 64 logarithm into a base 2 logarithm:
log%2864%2C+%2832%29%29+=+log%282%2C+%2832%29%29%2Flog%282%2C+%2864%29%29+=+5%2F6