sweetboy
Member
Registered: Apr 2002
Location: cali
Posts: 34
Status: Offline |
hahah..i'll just make up one..hehe..hm....
lets say you have Log(2)32
you set it equal to x so....log(2)32=x
and since b^y=x
2 to the power of x, equals 32
so 2^x=32
then you solve for x by simplyfying 32 in terms of 2..you can say that 32 = 2^5
so you have 2^x = 2^5
and since the bases are the same you can set the exponents equal to each other so x=5 and thats the answer...
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